2/x^3-x^2-x+1=$\frac{3}{1-x^2}$ – $\frac{1}{x+1}$ 3/(x+2)(x^2-1)=(x+2)(2x^2-2) 05/07/2021 Bởi Natalia 2/x^3-x^2-x+1=$\frac{3}{1-x^2}$ – $\frac{1}{x+1}$ 3/(x+2)(x^2-1)=(x+2)(2x^2-2)
`x^3-x^2-x+1= 3/( 1 – x^2) – 1/( x + 1)` `<=>x^3(x+1)(1-x)-x^2(1+x)(1-x)-x(1+x)(1-x)+(1+x)(1-x)=3+x-1` `<=>2x^3+x^4-x^5-2x^2-x+1-x-2=0` `<=>2x^3+x^4-x^5-2x^2-2x-1=0` `<=>x≈-1,3` Bình luận
Giải thích các bước giải: 3.$(x+2)(x^2-1)=(x+2)(2x^2-2)$ $\to (x+2)(2x^2-2)-(x+2)(x^2-1)=0$ $\to (x+2)(2x^2-2-x^2+1)=0$ $\to (x+2)(x^2-1)=0$ $\to (x+2)(x-1)(x+1)=0$ $\to x\in\{-2,1,-1\}$ Bình luận
`x^3-x^2-x+1= 3/( 1 – x^2) – 1/( x + 1)`
`<=>x^3(x+1)(1-x)-x^2(1+x)(1-x)-x(1+x)(1-x)+(1+x)(1-x)=3+x-1`
`<=>2x^3+x^4-x^5-2x^2-x+1-x-2=0`
`<=>2x^3+x^4-x^5-2x^2-2x-1=0`
`<=>x≈-1,3`
Giải thích các bước giải:
3.$(x+2)(x^2-1)=(x+2)(2x^2-2)$
$\to (x+2)(2x^2-2)-(x+2)(x^2-1)=0$
$\to (x+2)(2x^2-2-x^2+1)=0$
$\to (x+2)(x^2-1)=0$
$\to (x+2)(x-1)(x+1)=0$
$\to x\in\{-2,1,-1\}$