Question

giải phương trình sau:

a) \(\frac{x^2-6}{x}=x+\frac{3}{2}\)

b) \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)

Answer 1

\(a\)) \(ĐKXĐ:x\ne0\)

\(MTC:2x\)

Khi đó phương trình \(\Leftrightarrow2.\left(x^2-6\right)=2x^2+3x\)

\(\Leftrightarrow2x^2-12-2x^2-3x=0\Leftrightarrow3x=-12\)

\(\Leftrightarrow x=-4\left(nhận\right)\)

Vậy \(S=\left\{-4\right\}\)

Answer 2

\(b\)) \(ĐKXĐ:\left[{}\begin{matrix}x+1\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)

\(MTC:x\left(x+1\right)\)

Khi đó phương trình \(\Leftrightarrow x\left(x+3\right)+\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)\)

\(\Leftrightarrow x^2+3x+x^2-x-2=2x^2+2x\)

\(\Leftrightarrow2x^2+2x-2=2x^2+2x\)

\(\Leftrightarrow2x^2+2x-2-2x^2-2x=0\Leftrightarrow0x=-2\left(vônghiệm\right)\)

Vậy \(S=\varnothing\)