Question

Cho biểu thức: A= (\(\frac{x}{x-1}\)+ \(\frac{x}{x^2-1}\)) : (\(\frac{2}{x^2}\)-\(\frac{2-x^2}{x^3+x^2}\))

a) Rút gọn biểu thức A

b) Tìm x biết A=4

c) Tìm x để A<0

Answer 1

a) Ta có: A = \(\left(\frac{x}{x-1}+\frac{x}{x^2-1}\right):\left(\frac{2}{x^2}-\frac{2-x^2}{x^3+x^2}\right)\)

A = \(\left(\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2\left(x+1\right)}{x^2\left(x+1\right)}-\frac{2-x^2}{x^2\left(x+1\right)}\right)\)

A = \(\left(\frac{x^2+x+x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x+2-2+x^2}{x^2\left(x+1\right)}\right)\)

A = \(\left(\frac{x^2+2x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2+2x}{x^2\left(x+1\right)}\right)\)

A = \(\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x^2\left(x+1\right)}{x\left(x+2\right)}\)

A = \(\frac{x^2}{x+1}\)

b) ĐKXĐ: x \(\ne\)\(\pm\)1; x \(\ne\)0; x \(\ne\)-2

Ta có: A = 4

<=> \(\frac{x^2}{x+1}=4\)

<=> x² = 4(x + 1)

<=> x² - 4x - 4 = 0

<=>(x² - 4x + 4) - 8 = 0

<=> (x - 2)² = 8

<=> \(\orbr{\begin{cases}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\sqrt{2}+2\\x=2-2\sqrt{2}\end{cases}}\)(tm)

Vậy ...

c) Ta có: A < 0

<=> \(\frac{x^2}{x+1}< 0\)

Do x² \(\ge\)0 => x + 1 < 0

=> x < -1

Vậy để A < 0 thì x < -1 và x khác -2