We have: \(2^{2018}=\left(2^{1009}\right)^2=\left(2^{979}\cdot2^{30}\right)^2=\left(\left(2^{11}\right)^{89}\cdot2^{30}\right)^2\)

\(\equiv\left(48^{89}\cdot824\right)^2\left(mod1000\right)\)\(=\left(48^3\cdot\left(48^2\right)^{43}\cdot824\right)^2\)

\(\equiv\left(592\cdot304^{43}\cdot824\right)^2\left(mod1000\right)\)

\(\equiv\left(592\cdot\left(304^5\right)^8\cdot464\cdot824\right)^2\left(mod1000\right)\)

\(\equiv\left(912\cdot24^8\right)^2\left(mod1000\right)\)

\(\equiv\left(912\cdot776^2\right)^2\left(mod1000\right)\)

\(\equiv\left(912\cdot176\right)^2\left(mod1000\right)\)

\(\equiv512^2\left(mod1000\right)\equiv144\left(mod1000\right)\)

So the last three-digit numbers of \(M\) is \(144\)

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{c^2+b^2}{2}}+\sqrt{\dfrac{a^2+c^2}{2}}\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\sqrt{ab}\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\sqrt{ab}\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=Σ_{cyc}\dfrac{\left(a-b\right)^2}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\)

Or \(Σ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}-b\right)}{\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}+b\right)^2}}{\dfrac{1}{b}+\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}}\right)\right)\ge0\)

Done !!

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{a^2+c^2}{2}}\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\sqrt{ab}\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\sqrt{ab}\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=\dfrac{\left(a-b\right)^2}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\)

Or \(Σ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}-b\right)}{\dfrac{1}{b}+\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}}\right)\ge0\)

Done !!

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{c^2+a^2}{2}}\)

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\left(\sqrt{ab}\right)\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\left(\sqrt{ab}\right)\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=Σ_{cyc}\left(\dfrac{\left(a-b\right)^2}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\)

Or \(Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\geΣ_{cyc}\left(\dfrac{\left(a-b\right)^2}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}-b\right)}{b^2\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}\right)^2}}{\dfrac{1}{b}+\dfrac{1}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}}\right)\ge0\)

Done !!!

\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)

We have: \(x^2\ge0\forall x\Leftrightarrow x^2+1\ge1\forall x\)

\(\Leftrightarrow\dfrac{1}{x^2+1}\le1\forall x\)

\(\Leftrightarrow Q=\dfrac{3}{x^2+1}\le3\forall x\)

\("="\Leftrightarrow x=0\)

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\(\dfrac{x^2-z^2}{y+z}+\dfrac{y^2-x^2}{z+x}+\dfrac{z^2-y^2}{x+y}\ge0\)

\(\Leftrightarrow\dfrac{x^4+y^4+z^4-x^2y^2-y^2z^2-x^2z^2}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x^4-2x^2y^2+y^4\right)+\left(y^4-2y^2z^2+z^4\right)+\left(z^4-2x^2z^2+x^4\right)}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2+\left(y^2-z^2\right)^2+\left(z^2-x^2\right)^2}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\left(\text{*}\right)\)

The last inequality is obvious

\("="\Leftrightarrow x=y=z\)

By Cauchy-Schwarz's ineq:

\(VP=\dfrac{4a^2}{2ab+2ac}+\dfrac{4b^2}{2ab+2bc}+\dfrac{4c^2}{2ac+2bc}\)

\(\ge\dfrac{\left(2a+2b+2c\right)^2}{4\left(a+b+c\right)}=\dfrac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)}=a+b+c=VT\)

When \(a=b=c=1 \Rightarrow A=3\)

\(a=b=1\rightarrow a+\frac{1}{b}>1\)

Hey con lợn "ghuy-pu" kia bớt đú đi nhé, mạng VN đã quá thừa trẻ trâu vào rồi, lag lắm rồi 

if right-handed-sign is \(a^2b^2c^2\), you can assume it.

Stop Touching YourSelf, please =((

( ͡° ͜ʖ ͡°) 

#NOTE:\(e=equation\)

\(e\left(2\right)\Leftrightarrow y=3-3x\)

\(\Rightarrow e\left(1\right)\Leftrightarrow32x^3-48x^2+30x+\left(4\left(3-3x\right)-7\right)\sqrt{1-\left(3-3x\right)}=7\)

\(\Leftrightarrow32x^3-48x^2+30x+\left(5-12x\right)\sqrt{3x-2}=7\)

\(\Leftrightarrow32x^3-48x^2+10x+6+\left(5-12x\right)\sqrt{3x-2}-\left(-20x+13\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(4x-3\right)\left(4x+1\right)+\dfrac{\left(x-1\right)\left(4x-3\right)\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)\left(2\left(4x+1\right)+\dfrac{\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}\right)=0\)

See: \(2\left(4x+1\right)+\dfrac{\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}>0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=3-3x=3-3\cdot1=0\\y=3-3x=3-3\cdot\dfrac{3}{4}=\dfrac{3}{4}\end{matrix}\right.\)

We have: \(2006=2005+1=x+1\)

Hence \(B=x^{2005}-2006x^{2004}+2006x^{2003}-...+2006x+1\)

\(=x^{2005}-\left(x+1\right)x^{2004}+\left(x+1\right)x^{2003}-...+\left(x+1\right)x+1\)

\(=x^{2005}-x^{2005}-x^{2004}+x^{2014}+x^{2013}-...+x^2+x+1\)

\(=x+1=2005+1=2006\)

By Cauchy-Schwarz's inequality we have:

\(VT=\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\)

\(=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(y+z\right)}\)

\(\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{x+y}\right)+\dfrac{1}{4}\left(\dfrac{z}{x+z}+\dfrac{z}{z+y}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}\right)=\dfrac{3}{4}\)

When \(x=y=z\)

ok you can post your solution which you think smaller than I =))

A compact answer ? You need to know that this SOS method is one of method COMPACT because it doesn't loss of variables. 

Let \(A=\sqrt{6+\sqrt{6+\sqrt{6+....}}}>0\)

\(\Rightarrow A^2=6+\sqrt{6+\sqrt{6+...}}\)

\(\Rightarrow A^2=6+A\)\(\Rightarrow A^2-A-6=0\)

\(\Rightarrow\left(A+2\right)\left(A-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}A-3=0\\A+2=0\end{matrix}\right.\)\(\Rightarrow A=3\left(A>0\right)\)

Can you make the next question more clearly ?

By Cauchy-Schwarz's inequality:

\(L.H.S=\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\)

\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}\)

\(\ge\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}=R.H.S\)

DONE!

This pro will be solved by SOS method

\(\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\ge\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(\Leftrightarrow\dfrac{a^2b+a^2c-ab^2-ac^2}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab^2+b^2c-a^2b-bc^2}{\left(a^2+c^2\right)\left(a+c\right)}+\dfrac{bc^2+ac^2-a^2c-b^2c}{\left(a^2+b^2\right)\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab\left(b-a\right)+bc\left(c-b\right)}{\left(a^2+c^2\right)\left(a+c\right)}+\dfrac{bc\left(c-b\right)+ac\left(c-a\right)}{\left(a^2+b^2\right)\left(a+b\right)}\ge0\)

\(\LeftrightarrowΣ\left(\dfrac{ab\left(a-b\right)}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab\left(b-a\right)}{\left(a^2+c^2\right)\left(a+c\right)}\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)\left(\dfrac{\left(a^2+c^2\right)\left(a+c\right)-\left(b^2+c^2\right)\left(b+c\right)}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)\cdot\dfrac{\left(a-b\right)\left(a^2+b^2+c^2+ab+bc+ca\right)}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)^2\dfrac{a^2+b^2+c^2+ab+bc+ca}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\ge0\) *RIGHT!*

Exercise 2:

a)\(4x-x^2-5\)

\(=-x^2+4x-4-1\)

\(=-\left(x^2-4x+4\right)-1\)

\(=-\left(x-2\right)^2-1\le-1\)

When \(x=2\)

b)\(5-x^2-6x\)

\(=-x^2-6x-9+14\)

\(=-\left(x^2+6x+9\right)+14\)

\(=-\left(x+3\right)^2+14\le14\)

When \(x=-3\)

Search GTNN, GTLN @@

a)\(x^2-4x+1\)

\(=\left(x^2-4x+4\right)-3\)

\(=\left(x-2\right)^2-3\ge-3\)

When \(x=2\)

b)\(3x^2-6x-1\)

\(=3x^2-6x+3-4\)

\(=3\left(x^2-2x+1\right)-4\)

\(=3\left(x-1\right)^2-4\ge-4\)

When \(x=1\)

c)\(4x^2+4x-2\)

\(=4x^2+4x+1-3\)

\(=\left(2x+1\right)^2-3\ge-3\)

When \(x=-\dfrac{1}{2}\)

Use Cauchy-Schwarz's inequality we have:

\(L.H.S=\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\)

\(=\dfrac{1}{\dfrac{a+b}{ab}}+\dfrac{1}{\dfrac{b+c}{bc}}+\dfrac{1}{\dfrac{c+a}{ca}}\)

\(=\dfrac{ab}{a+b}+\dfrac{ac}{a+c}+\dfrac{bc}{b+c}\)

\(=a+b+c-\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)\)

\(\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)

\(=a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}=R.H.S\)

When \(a=b=c\)

Cảm ơn  ! Các bạn đã gõ ra những gì mà cả Topic này biết

\(75+12+43+25+88+50+57\)

\(=\left(75+25\right)+\left(12+88\right)+\left(43+57\right)+50\)

\(=100+100+100+50\)

\(=300+50=350\)

Use AM-GM's ineq:

\(\dfrac{b+c}{\sqrt{a}}+\dfrac{c+a}{\sqrt{b}}+\dfrac{a+b}{\sqrt{c}}\ge\dfrac{2\sqrt{bc}}{\sqrt{a}}+\dfrac{2\sqrt{ca}}{\sqrt{b}}+\dfrac{2\sqrt{ab}}{\sqrt{c}}\)

\(=\left(\sqrt{\dfrac{bc}{a}}+\sqrt{\dfrac{ca}{b}}\right)+\left(\sqrt{\dfrac{ca}{b}}+\sqrt{\dfrac{ab}{c}}\right)+\left(\sqrt{\dfrac{ab}{c}}+\sqrt{\dfrac{bc}{a}}\right)\)

\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)

\(\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{abc}\)\(=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)

Example:  they're \(\angle B=\angle M;\angle C=\angle Q;\angle A=\angle P\) but they aren't same. This is Similar triangles

i have a solution but it's ugly

If \(a=b=1\) and \(c=0\)   then we get a value  \(2+\frac{1}{\sqrt2}\)

We'll prove that it's a minimal value. Thus, we need to prove that

\(\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

WLOG \(c=\min\{a,b,c\}\). Hence:

\(\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0\)

Similar\(\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0\)

And \(\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0\)

Let \(\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)\)\(\Rightarrow xy=1\) and we have:

\(x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}\)

\(\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}\)

\(\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2\)

By Cauchy-Schwarz's ine we have: 

\(\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y\)

\(=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2\)

Thus, it's enough to prove that \(\sqrt{x^2+y^2}+\sqrt{2}\ge2\)

It's true by AM-GM \(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)

Use AM-GM's inequality we have:

\(\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{abc}{2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}}\)\(=\dfrac{abc}{8abc}=\dfrac{1}{8}\)

cube not square. this is a cube 

\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\) wrong too

\(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\) - it's wrong

Auto Ask Auto Answer ? i will report it :|

WLOG \(x\le y \le z\) we have:

\(\dfrac{7}{10}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}=\dfrac{3}{z}\)

\(\Rightarrow\dfrac{7}{10}\ge\dfrac{3}{z}\Rightarrow z\ge\dfrac{30}{7}\approx4,28\)

Because \(z\) is positive integer id est \(z\ge 5\)

Because \(x+y+z\) is smallest value should find \(z\) is smallest value

\(\Rightarrow z=5\Rightarrow x=3;y=6\) (wrong because \(x\le y \le z\))

\(\Rightarrow z=6\Rightarrow y=5\Rightarrow x=3\) (right because \(x\le y \le z\))

We only find 1 smallest value satisfy, should \(z=7;z=8;... \) wrong

Hence \(x+y+z=3+5+6=14\)

\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)\(\Leftrightarrow x+3=308\Rightarrow x=305\)