\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

`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`

(-3).8/8.6 rút gọn

\(=4x^2-9-20x-4x^2=-20x-9\)

(2x-3).(x+2)-(4x-2).(x-5)

Điều kiện: x ≠ 0; x ≠ 1 và x ≠ 3

chịu

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

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

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

\(=-22\)

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

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

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

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

\(=-22\)

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