Cíuu t cmayy oii,3 bài thoii

Giải thích các bước giải:

Bài 5:

ĐKXĐ: $x\ne 0, x\ne 1$

Ta có:

$(\dfrac{x+1}x+\dfrac{2+x-x^2}{x^2-x}):(\dfrac1{x-1}-\dfrac{x^2}{x^3-1})$

$=(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{2+x-x^2}{x\left(x-1\right)}):(\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x^2+x+1\right)})$

$=\dfrac{x+1}{x\left(x-1\right)}:\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}$

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

$=\dfrac{x^2+x+1}{x}$

Bài 6:

ĐKXĐ: $x\ne\pm3, x\ne -2$

Ta có:

$(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x})\cdot \dfrac{x+3}{x+2}$

$=(\dfrac{21}{(x+3)(x-3)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3})\cdot \dfrac{x+3}{x+2}$

$=(\dfrac{21}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)})\cdot \dfrac{x+3}{x+2}$

$=\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\cdot \dfrac{x+3}{x+2}$

$=\dfrac{3x+6}{(x+3)(x-3)}\cdot\dfrac{x+3}{x+2}$

$=\dfrac{3(x+2)}{(x+3)(x-3)}\cdot\dfrac{x+3}{x+2}$

$=\dfrac3{x-3}$

4.ĐKXĐ: $x\ne \pm2$

Ta có:

$(\dfrac{x+2}{x-2}+\dfrac{16}{4-x^2}-\dfrac{x-2}{x+2}):\dfrac{x+4}{x+2}$

$=(\dfrac{x+2}{x-2}-\dfrac{16}{x^2-4}-\dfrac{x-2}{x+2}):\dfrac{x+4}{x+2}$

$=(\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{16}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)})\cdot \dfrac{x+2}{x+4}$

$=\dfrac{\left(x+2\right)^2-16-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot \dfrac{x+2}{x+4}$

$=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}\cdot \dfrac{x+2}{x+4}$

$=\dfrac{8}{x+2}\cdot \dfrac{x+2}{x+4}$

$=\dfrac8{x+4}$