Câu hỏi của Hồng Minh

Question

Giúp mk nha

Giải các phương trình

1/ $\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}$

2/ \(\dfrac{1}{x^2-6x+8}+\dfrac{1}{x^2-10x+24}+\dfrac{1}{x^2-14x+48}=\dfrac{1}{9}\...

Mysterious Person · Accepted Answer

1) điều kiện xác định : $x\notin\left\{-1;-2;-3;-4\right\}$

ta có : $\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)$

$\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)$

$\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)$

$\Leftrightarrow18=\left(x+1\right)\left(x+4\right)$ ( vì điều kiện xác định )

$\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0$

$\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\x+7=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=2\x=-7\end{matrix}\right.\left(tmđk\right)$

vậy $x=2$ hoặc $x=-7$  mấy câu kia lm tương tự nha bnCâu trả lời: 1) điều kiện xác định : $x\notin\left\{-1;-2;-3;-4\right\}$