Bài 2: 

a) Ta có: \(A=\left(7x+5\right)^2+\left(3x-5\right)^2-\left(10-6x\right)\left(5+7x\right)\)

\(=\left(7x+5\right)^2+2\cdot\left(7x+5\right)\cdot\left(3x-5\right)+\left(3x-5\right)^2\)

\(=\left(7x+5+3x-5\right)^2\)

\(=\left(10x\right)^2=100x^2\)

Thay x=-2 vào A, ta được:

\(A=100\cdot\left(-2\right)^2=100\cdot4=400\)

b) Ta có: \(B=\left(2x+y\right)\left(y^2-2xy+4x^2\right)-8x\left(x-1\right)\left(x+1\right)\)

\(=8x^3+y^3-8x\left(x^2-1\right)\)

\(=8x^3+y^3-8x^3+8x\)

\(=8x+y^3\)

Thay x=-2 và y=3 vào B, ta được:

\(B=-2\cdot8+3^3=-16+27=11\)

Ai help mk vs

Bài 1: 

Ta có: \(A=x\left(4-x\right)\)

\(=4x-x^2\)

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

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

\(=-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x=2

Vậy: \(A_{max}=4\) khi x=2

Bài 3:

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy.......................

2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)

Vậy........................

3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy............................

4, 5 tương tự nhé bn!

bài 3

1 (x-1)(x+2)+5x-5=0

=>(x-1)(x+2)+(5x-5)=o

=>(x-1)(x+2)+5(x-1)=0

=>(x-1)(x+2+5)=0

=>(x-1)(x+7)=0

=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

vậy x=1 hoặc x=-7

2. (3x+5)(x-3)-6x-10=0

=>(3x+5)(x-3)-(6x+10)=0

=>(3x+5)(x-3)-2(3x+5)=0

=>(3x+5)(x-3-2)=0

=>(3x+5)(x-5)=0

=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)

\(a)\frac{1}{7}x-\frac{1}{2}x+\frac{5}{7}x=-\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{7}-\frac{1}{2}+\frac{5}{7}\right)x=-\frac{1}{2}\)

\(\Rightarrow\left(\frac{2}{14}-\frac{7}{14}+\frac{10}{14}\right)x=-\frac{1}{2}\)

\(\Rightarrow\frac{5}{14}x=-\frac{1}{2}\)

\(\Rightarrow x=-\frac{1}{2}:\frac{5}{14}\)

\(\Rightarrow x=-\frac{1}{2}.\frac{14}{5}\)

\(\Rightarrow x=-\frac{7}{5}\)

\(b)(\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{49.51})x=-\frac{1}{3}\)

\(\Rightarrow\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{49}-\frac{1}{51}\right)x=-\frac{1}{3}\)

\(\Rightarrow\left(\frac{1}{11}-\frac{1}{51}\right)x=-\frac{1}{3}\)

\(\Rightarrow\left(\frac{51}{561}-\frac{11}{561}\right)x=-\frac{1}{3}\)

\(\Rightarrow\frac{40}{561}x=-\frac{1}{3}\)

\(\Rightarrow x=-\frac{1}{3}:\frac{40}{561}\)

\(\Rightarrow x=-\frac{1}{3}.\frac{561}{40}\)

\(\Rightarrow x=-\frac{187}{40}\)

Chúc bạn học tốt !!! 

Lời giải:
a. 

PT $\Leftrightarrow -5x^2+15x-5+x+5x^2=x-2$
$\Leftrightarrow 16x-5=x-2$

$\Leftrightarrow 15x=3$

$\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}$

b.

PT $\Leftrightarrow -4x^2+20x+7x^2-28x-3x^2=12$

$\Leftrightarrow -8x=12$

$\Leftrightarrow x=\frac{-3}{2}$

a) |x - 5| - 2x = 3

| x - 5| = 3 + 2x

=> x - 5 = 3 + 2x hoặc x - 5 = -3 - 2x

=> -5 - 3 = 2x - x -5 + 3 = -2x - x

=> x = -8 -2 = -3x

=> x = 2/3

b) |2x - 1| + 3x = 1

|2x - 1| = 1 - 3x

=> 2x - 1 = 1 - 3x hoặc 2x - 1 = -1 + 3x

=> -1 - 1 = -3x - 2x -1 + 1 = 3x - 2x

=> -2 = -5x 0 = x

=> x = 2/5

c) | x - 5| = 3x - 2

=> x - 5 = 3x - 2 hoặc x - 5 = -3x + 2

=> -5 + 2 = 3x - x -5 - 2 = -3x - x

=> -3 = 2x -7 = -4x

=> x = -3/2 x = 7/4

d) |9 - 7x| = 5x - 3

=> 9 - 7x = 5x - 3 hoặc 9 - 7x = -5x + 3

=> 9 + 3 = 5x + 7x 9 - 3 = -5x + 7x

=> 12 = 12x 6 = 2x

=> x = 1 x = 3

d./9-7x/=5x-3 c./9-7x/=5x-3

\(\Rightarrow\) 9-7x=5x-3 \(\Rightarrow\)

9+3=5x+7x

12=(5+7)x

12=12x

vậy x=1

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