Question

giải phương trình

a) \(\frac{4x-8}{2x^2+1}=0\)

b)\(\frac{x^2-x-6}{x-3}=0\)

c)\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)

d)\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

Answer 1

a.ĐK: 2x²+1\(\ne0\) \(\forall x\)

Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x²+1 >0 với mọi x)

\(\Leftrightarrow x=2\) (TM)

Vậy ...

b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)

Để phương trình bằng 0 thì x²-x-6=0 (Vì x-3\(\ne0\))

\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)

Vậy ...

c. ĐK: x\(\ne\)2

\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)

\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)

\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)

\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)

Vậy ...

d. ĐK: \(x\ne\pm\frac{1}{3}\)

\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)

\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))

\(\Leftrightarrow x=-2\:\left(TM\right)\)

Vậy...