Câu hỏi của Trần Ích Bách

Question

Giải phương trình $\dfrac{15x}{x^2+3x-4}-1=12.\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)$

Otasaka Yu · Accepted Answer

$\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)$

ĐKXĐ: $\left\{{}\begin{matrix}x\ne-4\x\ne1\end{matrix}\right.$

$\Leftrightarrow\dfrac{15x-x^2-3x+4}{\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{3x-3+x+4}{3\left(x+4\right)\left(x-1\right)}\right)$

$\Leftrightarrow\dfrac{3(12x-x^2+4)}{3\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{4x+1}{3\left(x+4\right)\left(x-1\right)}\right)$

$\Leftrightarrow-x^2+12x+4=16x+4$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\x=-4\left(l\right)\end{matrix}\right.$

Vậy $S=\left\{0\right\}$Câu trả lời: $\dfrac{15x}{x^2+3x-4}-1=12\left(\dfrac{1}{x+4}+\dfrac{1}{3x-3}\right)$

ĐKXĐ: $\left\{{}\begin{matrix}x\ne-4\x\ne1\end{matrix}\right.$

$\Leftrightarrow\dfrac{15x-x^2-3x+4}{\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{3x-3+x+4}{3\left(x+4\right)\left(x-1\right)}\right)$

$\Leftrightarrow\dfrac{3(12x-x^2+4)}{3\left(x-1\right)\left(x+4\right)}=12\left(\dfrac{4x+1}{3\left(x+4\right)\left(x-1\right)}\right)$

$\Leftrightarrow-x^2+12x+4=16x+4$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\left(n\right)\x=-4\left(l\right)\end{matrix}\right.$

Vậy $S=\left\{0\right\}$

Otasaka Yu · Answer

qui đồng là xong lười làm hảCâu trả lời: qui đồng là xong lười làm hả

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