$\frac{x^2}{\left(x-1\right)^2}=3x^2+4x$ find x 29/07/2021 Bởi Kennedy $\frac{x^2}{\left(x-1\right)^2}=3x^2+4x$ find x
CHÚC BẠN HỌC TỐT!!! Trả lời: ĐKXĐ: $x\neq 1$ $\dfrac{x^2}{(x-1)^2}=3x^2+4x$ $⇔x.\bigg{(}3x+4-\dfrac{x}{(x-1)^2}\bigg{)}=0$ $⇔\left[ \begin{array}{l}x=0\\3x+4-\dfrac{x}{(x-1)^2}=0\,\,(1)\end{array} \right.$ $(1)⇔3x-2+6-\dfrac{x}{x^2-2x+1}=0$ $⇔3x-2+\dfrac{6x^2-13x+6}{x^2-2x+1}=0$ $⇔3x-2+\dfrac{(3x-2)(2x-3)}{x^2-2x+1}=0$ $⇔(3x-2).\bigg{(}1+\dfrac{2x-3}{x^2-2x+1}\bigg{)}=0$ $⇔\left[ \begin{array}{l}x=\dfrac{2}{3}\\1+\dfrac{2x-3}{x^2-2x+1}=0\,\,(2)\end{array} \right.\,\,\,(2)⇔\dfrac{x^2-2}{x^2+2x-1}=0$ $⇔x^2=2⇔x=±\sqrt{2}$ Vậy `S={0;\frac{2}{3};\sqrt{2};-\sqrt{2}}`. Bình luận
Đáp án: Giải thích các bước giải: $\frac{x²}{(x-1)²}$ = 3x²+4x⇔$\frac{x²}{(x-1)²}$ = $\frac{3x²(x-1)²}{(x-1)²}$ + $\frac{4x(x-1)²}{(x-1)²}$ ⇔$\frac{x²}{(x-1)²}$ =$\frac{3x²(x-1)²+4x(x-1)²}{(x-1)²}$ ⇔x²=(3x²+4x)(x-1)²⇔x²=(3x²+4x)(x²-2x+1)⇔x²=3x^4-6x³+3x²+4x³-8x²+4x⇔-3x^4+6x³-4x³+x²-3x²+8x²-4x=0⇔-3x^4+2x³+6x²-4x=0⇔(-3x^4+6x²)+(2x³-4x)=0⇔-3x²(x²-2)+2x(x²-2)=0⇔(-3x²+2x)(x²-2)=0⇔(-3x²+2x)(x²-2)=0⇔\(\left[ \begin{array}{l}-3x²+2x=0\\x²-2=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x(-3x+2)=0\\x²=-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=0\\-3x+2=0\end{array} \right.\) Hoặc: x=√2⇔\(\left[ \begin{array}{l}x=0\\x=$\frac{2}{3}$ \end{array} \right.\) Hoặc: x=√2Vậy ….. Bình luận
CHÚC BẠN HỌC TỐT!!!
Trả lời:
ĐKXĐ: $x\neq 1$
$\dfrac{x^2}{(x-1)^2}=3x^2+4x$
$⇔x.\bigg{(}3x+4-\dfrac{x}{(x-1)^2}\bigg{)}=0$
$⇔\left[ \begin{array}{l}x=0\\3x+4-\dfrac{x}{(x-1)^2}=0\,\,(1)\end{array} \right.$
$(1)⇔3x-2+6-\dfrac{x}{x^2-2x+1}=0$
$⇔3x-2+\dfrac{6x^2-13x+6}{x^2-2x+1}=0$
$⇔3x-2+\dfrac{(3x-2)(2x-3)}{x^2-2x+1}=0$
$⇔(3x-2).\bigg{(}1+\dfrac{2x-3}{x^2-2x+1}\bigg{)}=0$
$⇔\left[ \begin{array}{l}x=\dfrac{2}{3}\\1+\dfrac{2x-3}{x^2-2x+1}=0\,\,(2)\end{array} \right.\,\,\,(2)⇔\dfrac{x^2-2}{x^2+2x-1}=0$
$⇔x^2=2⇔x=±\sqrt{2}$
Vậy `S={0;\frac{2}{3};\sqrt{2};-\sqrt{2}}`.
Đáp án:
Giải thích các bước giải:
$\frac{x²}{(x-1)²}$ = 3x²+4x
⇔$\frac{x²}{(x-1)²}$ = $\frac{3x²(x-1)²}{(x-1)²}$ + $\frac{4x(x-1)²}{(x-1)²}$
⇔$\frac{x²}{(x-1)²}$ =$\frac{3x²(x-1)²+4x(x-1)²}{(x-1)²}$
⇔x²=(3x²+4x)(x-1)²
⇔x²=(3x²+4x)(x²-2x+1)
⇔x²=3x^4-6x³+3x²+4x³-8x²+4x
⇔-3x^4+6x³-4x³+x²-3x²+8x²-4x=0
⇔-3x^4+2x³+6x²-4x=0
⇔(-3x^4+6x²)+(2x³-4x)=0
⇔-3x²(x²-2)+2x(x²-2)=0
⇔(-3x²+2x)(x²-2)=0
⇔(-3x²+2x)(x²-2)=0
⇔\(\left[ \begin{array}{l}-3x²+2x=0\\x²-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x(-3x+2)=0\\x²=-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\-3x+2=0\end{array} \right.\)
Hoặc: x=√2
⇔\(\left[ \begin{array}{l}x=0\\x=$\frac{2}{3}$ \end{array} \right.\)
Hoặc: x=√2
Vậy …..