Giải pt a, 2. |x-1|=4 b, x-1/x+1 + 4/1-x^2 = 2.(x+1)/x-1 c,|x+1| + |x^2 + x-2 |= x^3 +1

Đáp án: a.$x\in\{-1,3\}$             b.$x=-5$                                 c.$x\in\{1,\sqrt2\}$

Giải thích các bước giải:

a.$2|x-1|=4\to |x-1|=2\to x-1=2\to x=3$ hoặc $x-1=-2\to x=-1$

b.ĐKXĐ : $x\ne \pm1$

$\dfrac{x-1}{x+1}+\dfrac{4}{1-x^2}=\dfrac{2(x+1)}{x-1}$

$\to\dfrac{x-1}{x+1}+\dfrac{4}{(1-x)(1+x)}=\dfrac{2(x+1)}{x-1}$

$\to\dfrac{x-1}{x+1}-\dfrac{4}{(x-1)(1+x)}=\dfrac{2(x+1)}{x-1}$

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

$\to \left(x-1\right)^2-4=2\left(x+1\right)^2$

$\to x^2-2x-3=2x^2+4x+2$

$\to -x^2-6x-5=0$

$\to -(x+5)(x+1)=0$
$\to x=-5, x\ne \pm1$

c.Ta có : $|x+1|+|x^2+x-2|\ge 0\to x^3+1\ge 0\to (x+1)(x^2-x+1)\ge 0\to x+1\ge 0\to x\ge -1$

$\to |x+1|=x+1$

$\to x+1+|x^2+x-2|=x^3+1$

$\to x^3-x=|x^2+x-2|$

$\to x^3-x=x^2+x-2\to x^3-x^2-2x+2=0\to \left(x-1\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0$

$\to x\in\{1,-\sqrt2,\sqrt2\}$

Hoặc $ x^3-x=-(x^2+x-2)$

$\to x^3+x^2-2=0$
$\to (x-1)(x^2+2x+2)=0$

$\to x=1$
$\to x\in\{1,-\sqrt2,\sqrt2\}$

Mà $x\ge -1\to x\in\{1,\sqrt2\}$