giải phương trình: $x^2$+$(\frac{x}{x-1})^2$ $= 8$ 13/10/2021 Bởi Maya giải phương trình: $x^2$+$(\frac{x}{x-1})^2$ $= 8$
Đáp án: $S=\{2\}$ Giải thích các bước giải: `ĐKXĐ: x\ne 1` `x^2+(\frac{x}{x-1})^2=8` `⇔\frac{x^2(x-1)^2}{(x-1)^2}+\frac{x^2}{(x-1)^2}=\frac{8(x-1)^2}{(x-1)^2}` `⇒x^2(x^2-2x+1)+x^2=8(x^2-2x+1)` `⇔x^4-2x^3+x^2+x^2=8x^2-16x+8` `⇔x^4-2x^3+2x^2=8x^2-16x+8` `⇔x^4-2x^3-6x^2+16x-8=0` `⇔x^4+2x^3-2x^2-4x^3-8x^2+8x+4x^2+8x-8=0` `⇔x^2(x^2+2x-2)-4x(x^2+2x-2)+4(x^2+2x-2)=0` `⇔(x^2-4x+4)(x^2+2x-2)=0` `⇔(x-2)^2(x^2+2x-2)=0` \(⇔\left[ \begin{array}{l}(x-2)^2=0\\x^2+2x-2=0\end{array} \right.\) \(⇔\left[ \begin{array}{l}x=2(tm)\\x^2+2x-2\end{array} \right.\)(vô nghiệm) Vậy $S=\{2\}$ Bình luận
x^2+(x/x-1)^2 = 8 x^2 + x^2/(x-1)^2 = 8 x^2(x-1)^2+x^2 = 8(x-1)^2 x^2(x^2-2x+1)+x^2=8(x^2-2x+1) x^4-2x^3+x^2+x^2=8x^2-16x+8 x^4-2x^3+x^2+x^2-8x^2+16x-8=0 x^4-2x^3-6x^2+16x-8=0 (????−2)^2(????^2+2????−2) = 0 => x-2=0 hoặc x^2+2x-2=0 x-2=0 =>x=2 x^2+2x-2=0 => x≈-2,73 vậy S={2 ; -2,72} Bình luận
Đáp án:
$S=\{2\}$
Giải thích các bước giải:
`ĐKXĐ: x\ne 1`
`x^2+(\frac{x}{x-1})^2=8`
`⇔\frac{x^2(x-1)^2}{(x-1)^2}+\frac{x^2}{(x-1)^2}=\frac{8(x-1)^2}{(x-1)^2}`
`⇒x^2(x^2-2x+1)+x^2=8(x^2-2x+1)`
`⇔x^4-2x^3+x^2+x^2=8x^2-16x+8`
`⇔x^4-2x^3+2x^2=8x^2-16x+8`
`⇔x^4-2x^3-6x^2+16x-8=0`
`⇔x^4+2x^3-2x^2-4x^3-8x^2+8x+4x^2+8x-8=0`
`⇔x^2(x^2+2x-2)-4x(x^2+2x-2)+4(x^2+2x-2)=0`
`⇔(x^2-4x+4)(x^2+2x-2)=0`
`⇔(x-2)^2(x^2+2x-2)=0`
\(⇔\left[ \begin{array}{l}(x-2)^2=0\\x^2+2x-2=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=2(tm)\\x^2+2x-2\end{array} \right.\)(vô nghiệm)
Vậy $S=\{2\}$
x^2+(x/x-1)^2 = 8
x^2 + x^2/(x-1)^2 = 8
x^2(x-1)^2+x^2 = 8(x-1)^2
x^2(x^2-2x+1)+x^2=8(x^2-2x+1)
x^4-2x^3+x^2+x^2=8x^2-16x+8
x^4-2x^3+x^2+x^2-8x^2+16x-8=0
x^4-2x^3-6x^2+16x-8=0
(????−2)^2(????^2+2????−2) = 0
=> x-2=0 hoặc x^2+2x-2=0
x-2=0 =>x=2
x^2+2x-2=0 => x≈-2,73
vậy S={2 ; -2,72}