Câu hỏi của Bùi Lê Trâm Anh

Question

giai phương trình:
a) $\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}$
b) $3x^2+2x-1=0$
c) $\frac{x-3}{x-2}+\frac{x-2}{4}=3\frac{1}{5}$

Y · Accepted Answer

a) $\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)$

$\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}$

$\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0$

$\Leftrightarrow x+100=0$    ( do $\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0$ )

$\Leftrightarrow x=-100$

b) $3x^2+2x-1=0$

$\Leftrightarrow3x^2+3x-x-1=0$

$\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0$

$\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}3x-1=0\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\x=-1\end{matrix}\right.$Câu trả lời: a) $\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)$