Đáp án:
`a.`Rút gọn:
`P=(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}):\frac{x^2-3x}{2x^2-x^3}`
`= (\frac{2+x}{-(x-2)}-\frac{4x^2}{(x-2)(x+2)}-\frac{2-x}{x+2})xx\frac{2x^2-x^3}{x^2-3x}`
`= (-\frac{2+x}{x-2}-\frac{4x^2}{(x-2)(x+2)}-\frac{2-x}{x+2})xx\frac{2x^2-x^3}{x^2-3x}`
`= -\frac{(x+2)^2+4x^2+(x-2)(2-x)}{(x-2)(x+2)}xx\frac{x^2(-(x-2))}{x(x-3)}`
`= -\frac{(x+2)^2+4x^2+(x-2)(-(x-2))}{(x-2)(x+2)}xx\frac{x^2(-(x-2))}{x(x-3)}`
`= -\frac{(x+2)^2+4x^2-(x-2)^2}{(x-2)(x+2)}xx\frac{x^2(-(x-2))}{x(x-3)}`
`= -\frac{8x+4x^2}{(x-2)(x+2)}xx\frac{x^2(-(x-2))}{x(x-3)}`
`= -\frac{4x(2+x)}{(x-2)(x+2)}xx\frac{x^2(-(x-2))}{x(x-3)}`
`= -\frac{4x}{x-2}xx\frac{x^2(-(x-2))}{x(x-3)}`
`= -4xx\frac{x^2 xx(-1)}{x-3}`
`= 4xx\frac{x^2}{x-3}`
`= \frac{4x^2}{x-3}`
`b.`Tìm `x`:
Để `P=5` thì `\frac{4x^2}{x-3}=8`
`<=> 4x^2=8(x-3)`
`<=> 4x^2=8x-24`
`<=> 4x^2-8x+24=0`
`<=> x^2-2x+6=0`
`<=> x^2-2x=-6`
`<=> x^2-2x+1=-6+1`
`<=> (x-1)^2=-5` (`x\in` $\varnothing$ `, (x-1)^2 >=0 \forall x`)
`=> x\in`$\varnothing$.
