\(\dfrac{3-x}{100}-1=\dfrac{2-x}{101}+\dfrac{1-x}{102}\)

\(\Leftrightarrow\dfrac{3-x}{100}+1=\dfrac{2-x}{101}+1+\dfrac{1-x}{102}+1\)

\(\Leftrightarrow\dfrac{103-x}{100}-\dfrac{103-x}{101}-\dfrac{103-x}{102}=0\)

\(\Leftrightarrow\left(103-x\right)\left(\dfrac{1}{100}-\dfrac{1}{101}-\dfrac{1}{102}\right)=0\)

We have: \(\dfrac{1}{100}-\dfrac{1}{101}-\dfrac{1}{102}\ne0\)

\(\Rightarrow103-x=0\Rightarrow x=103\)

There are empty seats in the train is:

\(60\cdot40\%=\dfrac{60\cdot40}{100}=24\) (seats)

Hence there are 24 empty seats in the train

We have: \(ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow\left(a+b\right)^2\ge4\Rightarrow a+b\ge2\left(a,b>0\right)\)

And \(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge ab\left(a+b\right)\ge2\)

By AM-GM's inequality we have:

\(\dfrac{a^3}{b+1}=a^3-\dfrac{a^3b}{b+1}\ge a^3-\dfrac{a^3b}{2\sqrt{b}}=a^3-\dfrac{a^3\sqrt{b}}{2}\)

Similar we have: \(\dfrac{b^3}{a+1}\ge b^3-\dfrac{b^3\sqrt{a}}{2}\)

\(\Rightarrow L.H.S\ge a^3+b^3-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\)

Need to prove: \(L.H.S\ge2-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge1=R.H.S\)

Use AM-GM's inequality:\(\sqrt{b}\le\dfrac{b+1}{2}\Rightarrow a^3\sqrt{b}\le\dfrac{a^3b+a^3}{2}\)

\(\Rightarrow\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\le\dfrac{\dfrac{a^3b+a^3}{2}+\dfrac{ab^3+b^3}{2}}{2}=\dfrac{a^3b+ab^3+a^3+b^3}{4}\)

\(\Rightarrow2-\dfrac{a^3\sqrt{b}+b^3\sqrt{a}}{2}\ge2-\dfrac{a^3b+ab^3+a^3+b^3}{4}\)

Hence \(2-\dfrac{a^3b+ab^3+a^3+b^3}{4}\ge1\Leftrightarrow\dfrac{a^3b+ab^3+a^3+b^3}{4}\ge1\)

\(\Leftrightarrow a^3b+ab^3+a^3+b^3\ge4\)\(\Leftrightarrow a^3b+ab^3\ge2\) \(\left(a^3+b^3\ge2\right)\)

\(\Leftrightarrow\left(ab\right)^2\left(a+b\right)\ge2\). Right because \(ab=1;a+b\ge2\)

Done !!

b)\(A=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+....+27\cdot28\cdot29\cdot30\)

\(5A=1\cdot2\cdot3\cdot4\cdot5+2\cdot3\cdot4\cdot5\cdot\left(6-1\right)+....+27\cdot28\cdot29\cdot30\cdot\left(31-26\right)\)

\(5A=1\cdot2\cdot3\cdot4\cdot5+2\cdot3\cdot4\cdot5\cdot6-1\cdot2\cdot3\cdot4\cdot5+....+27\cdot28\cdot29\cdot30\cdot31-26\cdot27\cdot28\cdot29\cdot30\)

\(5A=27\cdot28\cdot29\cdot30\cdot31\Rightarrow A=\dfrac{27\cdot28\cdot29\cdot30\cdot31}{5}=4077864\)

WaTerFall in Dictionary haven't word "Đặt"

1,\(x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

2,\(5x\left(x-2\right)-x+2=0\)

\(\Leftrightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2\end{matrix}\right.\)

File downloaded from computer without link

We have 4 even numbers and 4 odd numbers, and :

- Odd plus odd is even

- Odd plus even is odd

There are 8 out of 14 pairs that are odd plus even. Therefore, the probability of an odd sum \(\dfrac{8}{14}=0,57\) or \(57\%\)

My try: WLOG \(a\ge b\ge c\) hence we have: \(a^{m-n}\ge b^{m-n}\ge c^{m-n}\)

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

By Chebyshev's inequality we have: 

\(\text{∑}\dfrac{a^m}{b^n+c^n}=\text{∑}a^{m-n}\cdot\dfrac{a^n}{b^n+c^n}\)

\(\ge\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{3}\cdot\text{∑}\dfrac{a^n}{b^n+c^n}\)

\(\ge\text{∑}\dfrac{a^{m-n}+b^{m-n}+c^{m-n}}{2}\)

We're done when \(a=b=c\)

P/s: This's discuss not A⁴ (Auto Ask Auto Answer)

a)\(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(=\left(x^2+5xy+6y^2\right)\left(x^2+5xy+4y^2\right)+y^4\)

Let \(x^2+5xy=a;y^2=b\) we have;

\(=\left(a+6b\right)\left(a+4b\right)+b^2\)

\(=a^2+10ab+24b^2+b^2\)

\(=a^2+10ab+25b^2=\left(a+5b\right)^2\)

\(=\left(x^2+5xy+5y^2\right)^2\)

b)\(B=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

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

Let \(a=x^2+3x\) we have:

\(=a\left(a+2\right)+1=a^2+2a+1\)

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

Done !!

a)\(A=x^2+3x+7\)

\(=x^2+3x+\dfrac{9}{4}+\dfrac{19}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

When \(x=-\dfrac{3}{2}\)

Từ \(a+b+c=0\Rightarrow b+c=-a\)

\(\Rightarrow\left(b+c\right)^2=\left(-a\right)^2\Rightarrow b^2+2bc+c^2=a^2\)

\(\Rightarrow b^2+c^2-a^2=-2bc\)\(\Rightarrow\left(b^2+c^2-a^2\right)^2=\left(-2bc\right)^2\)

\(\Rightarrow b^4+c^4+a^4+2b^2c^2-2a^2b^2-2a^2c^2=4b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)

\(\Rightarrow2(a^4+b^4+c^4)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\)

\(\Rightarrow2(a^4+b^4+c^4)=(a^2+b^2+c^2)^2\)

a,b>0; c<0 ? or c=0 ? seem to have a typo

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

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(2x-4\right)^2}=3\)

\(\Leftrightarrow\sqrt{x^2-2x+1}+\sqrt{4x^2-16x+16}=3\)

\(\Leftrightarrow\sqrt{x^2-2x+1}-\left(\dfrac{2}{3}x-\dfrac{1}{9}\right)+\sqrt{4x^2-16x+16}-\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)=0\)

\(\Leftrightarrow\dfrac{x^2-2x+1-\left(\dfrac{2}{3}x-\dfrac{1}{9}\right)^2}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{4x^2-16x+16-\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)^2}{\sqrt{4x^2-16x+16}+\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)}=0\)

\(\Leftrightarrow\dfrac{\dfrac{45x^2-150x+80}{81}}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{\dfrac{288x^2-960x+512}{81}}{\sqrt{4x^2-16x+16}+\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)}=0\)

\(\Leftrightarrow\dfrac{\dfrac{5\left(3x-8\right)\left(3x-2\right)}{81}}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{\dfrac{32\left(3x-8\right)\left(3x-2\right)}{81}}{\sqrt{4x^2-16x+16}+\left(-\dfrac{2}{3}x+\dfrac{28}{9}\right)}=0\)

\(\Leftrightarrow\dfrac{\left(3x-8\right)\left(3x-2\right)}{81}\left(\dfrac{5}{\sqrt{x^2-2x+1}+\dfrac{2}{3}x-\dfrac{1}{9}}+\dfrac{32}{\sqrt{4x^2-16x+16}-\dfrac{2}{3}x+\dfrac{28}{9}}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-8=0\\3x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

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

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

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

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

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

a)\(x^2+3x+7\)

\(=x^2+3x+\dfrac{9}{4}+\dfrac{19}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

When \(x=-\dfrac{3}{2}\)

b)\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=\left(x^2-6xy+4x+9y^2-12y+4\right)+\left(x^2-10x+25\right)+1975\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1975\ge1975\)

When \(x=5;y=\dfrac{7}{3}\)

surf before ask Câu hỏi của Mai Nguyễn Bảo Ngọc - Toán lớp 8 | Học trực tuyến

From \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+3ab(a+b)+b^3+c^3-3abc-3ab(a+b)=0\)

\(\Rightarrow (a+b)^3+c^3-3ab(a+b+c)=0\)

\(\Rightarrow (a+b+c)(a^2+2ab+b^2-ab-ac+c^2)-3ab(a+b+c)=0\)

\(\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\)

It's right because \(a+b+c=0\)

I don't care. You need create a separate list to be somewhere in the web. This is spam, you're spammer

a)\(\left(x-1\right)\left(3-x\right)>0\)

\(\Leftrightarrow-\left(x-1\right)\left(x-3\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)< 0\)

\(\Rightarrow\)\(x-1;x-3\) are the numbers opposite the sign

We have: \(x-1>x-3\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}x-1>0\\x-3< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\)

b)\(xy=x+y\)

\(\Rightarrow xy-x-y+1=1\)

\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(;\left\{{}\begin{matrix}x-1=-1\\y-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y=2\\x=y=0\end{matrix}\right.\)