giải pt sau a) (x-2) : (x+3)-4x=(x-1)^2 06/11/2021 Bởi Sarah giải pt sau a) (x-2) : (x+3)-4x=(x-1)^2
$ĐKXĐ:x \neq -3$ $\to \dfrac{x-2}{x+3}-4x=(x-1)^2$ $\to x-2-4x(x+3)=(x^2-2x+1)(x+3)$ $\to x-2-4x^2-12x=x^3+3x^2-2x^2-6x+x+3$ $\to -4x^2-11x-2=x^3+x^2-5x+3$ $\to x^3+5x^2+6x+5=0$ $\to x=-3,757278921(TM)$ Bình luận
Đáp án: $\begin{array}{l}Dkxd:x \ne – 3\\\dfrac{{x – 2}}{{x + 3}} – 4x = {\left( {x – 1} \right)^2}\\ \Rightarrow \dfrac{{x – 2 – 4x\left( {x + 3} \right)}}{{x + 3}} = \dfrac{{{{\left( {x – 1} \right)}^2}\left( {x + 3} \right)}}{{x + 3}}\\ \Rightarrow x – 2 – 4{x^2} – 12x = \left( {{x^2} – 2x + 1} \right)\left( {x + 3} \right)\\ \Rightarrow – 4{x^2} – 11x – 2 = {x^3} + {x^2} – 5x + 3\\ \Rightarrow {x^3} + 5{x^2} + 6x + 5 = 0\\ \Rightarrow x = – 3,76\left( {tm} \right)\\Vậy\,x = – 3,76\end{array}$ Bình luận
$ĐKXĐ:x \neq -3$
$\to \dfrac{x-2}{x+3}-4x=(x-1)^2$
$\to x-2-4x(x+3)=(x^2-2x+1)(x+3)$
$\to x-2-4x^2-12x=x^3+3x^2-2x^2-6x+x+3$
$\to -4x^2-11x-2=x^3+x^2-5x+3$
$\to x^3+5x^2+6x+5=0$
$\to x=-3,757278921(TM)$
Đáp án:
$\begin{array}{l}
Dkxd:x \ne – 3\\
\dfrac{{x – 2}}{{x + 3}} – 4x = {\left( {x – 1} \right)^2}\\
\Rightarrow \dfrac{{x – 2 – 4x\left( {x + 3} \right)}}{{x + 3}} = \dfrac{{{{\left( {x – 1} \right)}^2}\left( {x + 3} \right)}}{{x + 3}}\\
\Rightarrow x – 2 – 4{x^2} – 12x = \left( {{x^2} – 2x + 1} \right)\left( {x + 3} \right)\\
\Rightarrow – 4{x^2} – 11x – 2 = {x^3} + {x^2} – 5x + 3\\
\Rightarrow {x^3} + 5{x^2} + 6x + 5 = 0\\
\Rightarrow x = – 3,76\left( {tm} \right)\\
Vậy\,x = – 3,76
\end{array}$