`Answer:`

`a)`

`A=5(x+1)^2-3(x-3)^2-4(x^2-4)`

`=>A=5(x^2+2x+1)-3(x^2-6x+9)-4x^2+16`

`=>A=5x^2+10x+5-3x^2+18x-27-4x^2+16`

`=>A=(5x^2-3x^2-4x^2)+(10x+18x)+(5-27+16)`

`=>A=-2x^2+28x-6`

`b)`

`B=5(x+1)^2-3(x-3)^2-4(x+2)(x-2)`

`=2x(3x+5)-3(3x+5)-2x(x^2-4x+4)-[(2x)^2-3^2]`

`=6x^2+10x-9x-15-2x^3+8x^2-8x-4x^2+9`

`=(6x^2-4x^2+8x^2)-2x^3+(10x-9x-8x)+(-15+9)`

Thay `x=-7` vào ta được:

`B=10(-7)^2-2(-7)^3-7(-7)-6`

`=>B=10.49-2(-343)+49-6`

`=>B=490+686+49-6`

`=>B=1219`

\(B=\left(3x-1\right)^2-\left(x+7\right)^2-2\left(2x-5\right)\left(2x+5\right)\)

\(=9x^2-6x+1-\left(x^2+14x+49\right)-2\left(4x^2-25\right)\)

\(=9x^2-6x+1-x^2-14x-49-8x^2+50\)

\(=-20x+2\)

Thay x=1/5 vào B, ta được:

\(B=-20\cdot\dfrac{1}{5}+2=-4+2=-2\)

Bạn gõ cái này nhé 

a: \(A=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{2x^2-2x-50}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\dfrac{x^2-10x+25}{2x\left(x-5\right)}=\dfrac{x-5}{2x}\)

b: Để A=1/3 thì x-5/2x=1/3

=>3x-15=2x

=>x=15

\(-2\sqrt{x^2}+2x+8\)

\(=-2\left|x\right|+2x+8\)

\(=-2x+2x+8\)

\(=8\)

ko biết

\(2x^3\left(x^2-5\right)+\left(-2x^3+4x\right)+\left(6+x\right)x^2\)

\(=2x^5-10x^3-2x^3+4x+6x^2+x^3=2x^5-9x^3+6x^2+4x\)

ko biet

mai minh 

học bài

này rùi bn 

ráng đợi thêm

2 ngày nữa nhé

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

\(=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x-7\)

\(=2x^2-7x-15-2x^2+6x+x-7\)

\(=-22\)

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

\(=2x^2+3x-10x-15-\left(2x^2-6\right)+x-7\)

\(=2x^2+3x-10x-15-2x^2+6x+x-7\)

\(=\left(2x^2-2x^2\right)+\left(3x-10x+6x+x\right)-15-7\)

\(=-22\)

\(\hept{\begin{cases}\frac{x^2}{2}-\frac{1}{2}=y\left(1-x-y\right)\\x\left(3x+2y-2\right)=y^2-y+5\end{cases}}\)

\(a,M=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\\ b,N=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)

\(N=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\dfrac{1}{x+\sqrt{5}}\)