a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)

\(=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{x+2}{\sqrt{x}}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\)

\(=\dfrac{x+2}{\sqrt{x}}\)

a. x² - 2x

⇔ x(x - 2)

b. 3x - 6y

⇔ 3(x - 2y)

c. 5(x + 3y) - 15x(x + 3y)

⇔ (5 - 15x)(x + 3y)

d. 3(x - y) - 5x(y - x)

⇔ 3(x - y) + 5x(x - y)

⇔ (3 + 5x)(x - y)

2B: 

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

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

c: \(xy\left(x-y\right)-2x+2y=\left(x-y\right)\left(xy-2\right)\)

d: \(y\left(x-y\right)^2+x\left(y-x\right)^2=\left(x-y\right)^2\cdot\left(x+y\right)\)

#include <bits/stdc++.h>

using namespace std;

double a,b;

int main()

{

cin>>a>>b;

cout<<(a+b)*2;

return 0;

}

Ta có |5x-5012| = |-5x+5012| >_ -5x +5012.Dấu "=" khi -5x+5012>0

         | 5x+300|>_ 5x+300.Dấu "=" khi 5x+300>0

=> |-5x+5012| + |5x+300| >_ -5x+5012 +5x + 300

=> A >_ 5312

Dấu "=" khi -5x+5012>0 => x<5012/5

                   5x+300> 0   => x>-60

Vậy Min A = 5312 khi -60<x<5012/5

\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=-1\end{matrix}\right.\)

8: \(=\dfrac{x_1^2+x_2^2}{\left(x_1\cdot x_2\right)^2}=\dfrac{2}{1}=2\)

9: \(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=2^3-3\cdot\left(-1\right)\cdot2=8+6=14\)

16: \(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}=\sqrt{2^2-4\cdot\left(-1\right)}=\sqrt{4+4}=2\sqrt{2}\)