Put \(\left\{{}\begin{matrix}x=b+c-a>0\\y=c+a-b>0\\z=a+b-c>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\\c=\dfrac{x+y}{2}\end{matrix}\right.\)

Instead we have:A=\(\dfrac{y+z}{2x}+\dfrac{z+x}{2y}+\dfrac{x+y}{2z}=\dfrac{1}{2}\left(\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{z}{y}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{z}\right)\)

Applying Cauchy's theorem for 2 numbers not negative we have:

A\(=\dfrac{1}{2}\left(\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{z}{x}+\dfrac{x}{z}+\dfrac{y}{z}+\dfrac{z}{y}\right)\)\(\ge\dfrac{1}{2}\left(2\sqrt{\dfrac{y}{x}.\dfrac{x}{y}}+2\sqrt{\dfrac{z}{x}.\dfrac{x}{z}}+2\sqrt{\dfrac{y}{z}.\dfrac{z}{y}}\right)\)

\(=\dfrac{1}{2}\left(2+2+2\right)=\dfrac{1}{2}.6=3\)

\(\Rightarrow\)The problem must prove and A=3 when x=y=z

Dao Trong Luan:Why? tan(a)=\(\dfrac{1}{2}\) can \(\Rightarrow a=26,5605118\)

This is you use calculator for calculate

For a= 2 . 3 ................ 1000001 .Because a>2 so a+2\(⋮2\)  so a+2 is a composite numbers

And a+3\(⋮3\) so a+3 is a composite numbers

............

And a+1000001\(⋮1000001\) so a+1000001 is a composite numbers

So a+2;a+3;..........;a+1000001 are 1000000 consecutive composite numbers

Method:Suppose

We have:\(1-\dfrac{1}{1+2+......+n}=1-\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=1-\dfrac{2}{n\left(n+1\right)}=\dfrac{n\left(n+1\right)-2}{n\left(n+1\right)}=\dfrac{n^2+n-2}{n\left(n+1\right)}\)

\(=\dfrac{n^2-n+2n-2}{n\left(n+1\right)}=\dfrac{n\left(n-1\right)+2\left(n-1\right)}{n\left(n+1\right)}=\dfrac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

Instead in the expression we have:

A=\(\dfrac{4.1}{2.3}+\dfrac{5.2}{3.4}+.......+\dfrac{1988.1985}{1986.1987}\)

\(=\dfrac{1.2....1985}{3.4....1987}.\dfrac{4.5.....1988}{2.3.....1986}\)

\(=\dfrac{1.2}{1986.1987}.\dfrac{1987.1988}{2.3}=\dfrac{1988}{3.1986}=\dfrac{1988}{5958}=\dfrac{994}{2979}\)

P/s:A is the expression

We have:\(\dfrac{y}{x^3-1}-\dfrac{x}{y^3-1}=\dfrac{y}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x}{\left(y-1\right)\left(y^2+y+1\right)}\)

\(=\dfrac{y}{\left(x-x-y\right)\left(x^2+x+1\right)}-\dfrac{x}{\left(y-x-y\right)\left(y^2+y+1\right)}\)

\(=\dfrac{y}{-y\left(x^2+x+1\right)}+\dfrac{-x}{-x\left(y^2+y+1\right)}=\dfrac{-1}{x^2+x+1}+\dfrac{1}{y^2+y+1}\)

\(=\dfrac{-y^2-y-1+x^2+x+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\)

\(=\dfrac{x^2-y^2+x-y}{x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)+x-y}{x^2y^2+xy\left(x+y\right)+x^2+y^2+x+y+xy+1}\)

\(=\dfrac{2x-2y}{x^2y^2+xy+1-2xy+1+xy+1}\)

\(=\dfrac{2\left(x-y\right)}{x^2y^2+3}\)

We have:n⁵-n=n(n⁴-1)=n(n²-1)(n²+1)=(n-1).n.(n+1).(n²-4+5)

=(n-2)(n-1)n(n+1)(n+2) + 5(n-1)n(n+1)\(⋮10\)

So n⁵\(\equiv n\)  (mod 10)

So the final digit of the numbers are in the form of n and n⁵ are in the same

\(x+2< \dfrac{5}{x-2}\Leftrightarrow\dfrac{5}{x-2}-\left(x+2\right)>0\)

\(\Rightarrow\dfrac{5-\left(x+2\right)\left(x-2\right)}{x-2}>0\Rightarrow\dfrac{5-\left(x^2-4\right)}{x-2}>0\)

\(\Rightarrow\dfrac{9-x^2}{x-2}>0\Rightarrow\left[{}\begin{matrix}9-x^2< 0;x-2< 0\\9-x^2>0;x>0\end{matrix}\right.\)

With 9-x²<0 and x-2<0 so x²>9 \(\Rightarrow\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\) .Because x-2<0 so x<2 so x<-3

\(\Rightarrow-10\le x< -3\)

With 9-x²>0 and x-2>0 so 9>x² \(\Rightarrow-3< x< 3\).Because x-2>0 so x>2 so there is no x satisfies the inequality

So the answer is:\(-10\le x< 3\)

We have:5\(\Delta\)3=\(\dfrac{5-3}{2.3+6}=\dfrac{2}{12}=\dfrac{1}{6}\)

So \(3\Delta\left(5\Delta3\right)=3\Delta\dfrac{1}{6}=\dfrac{3-\dfrac{1}{6}}{2.\dfrac{1}{6}+6}=\dfrac{17}{38}\)

Put \(x\) is the number of  males;n is the number of male students must be moved to another school

We have:\(\left\{{}\begin{matrix}\dfrac{x}{2000}.100=70\\\dfrac{x-n}{2000-n}.100=60\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x=1400\\x-n=0,6\left(2000-n\right)\end{matrix}\right.\)\(\Rightarrow1400-n=1200-0,6n\)

\(\Rightarrow0,4n=200\Rightarrow n=200:0,4=500\)

So the answer is:500 male students

So the units digit of n is:4

We have:\(\dfrac{3.3.3.3.3}{N}=\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{3}\Rightarrow\dfrac{3^5}{N}=\dfrac{1}{2}.\dfrac{1}{3^2}\)

\(\Rightarrow N=\dfrac{3^5}{\dfrac{1}{2}.\dfrac{1}{3^2}}=3^5.3^2.2=3^7.2=4374\)

 The answer is:

 1000:100 x 6=60($)

We have:\(\sqrt{150}+\sqrt{75}=5\sqrt{6}+5\sqrt{3}=5\left(\sqrt{6}+\sqrt{3}\right)=5.4,18154055=20,9077027\)

So the answer is:20

We have:\(g\left(8\right)=f\left(8\right)-3.8-7=f\left(8\right)-31\)

  \(f\left(8\right)=5.8=40\Rightarrow g\left(8\right)=40-31=9\)

The answer is:

 \(\dfrac{5.00}{5.00+5.00+5.00+5.00}=\dfrac{1}{4}\)

The volume of the smaller can is:

  \(\left(\dfrac{6}{2}\right)^2.3,14.20=565,2\)(cm³)

The volume of the larger volume is:

 \(\left(\dfrac{8}{2}\right)^2.3,14.17=854,08\)(cm³)

So the answer is:\(\dfrac{565,2}{854,08}=\dfrac{45}{68}\)

Because AP=2PB and AB=10-(-2)=12 so AP\(=\dfrac{AB}{1+2}=\dfrac{12}{3}=4\)

So the answer is:(6,7)

The answer is:

 \(\dfrac{92.12+10.87}{12+10}=\dfrac{987}{11}\approx89,72\)

The area of the circle is:

    \(\left(7:2\right)^2.3,14=38,465\)

The area of the square is:

   \(\left(\sqrt{\dfrac{7^2}{2}}\right)^2=\dfrac{49}{2}\)

So the answer is:\(\left|38,465-\dfrac{49}{2}\right|=\left|13,965\right|\approx13,97\)