Question

Giai phương trình sau :

a) \(\dfrac{4x-17}{2x^2+1}\) = 0

b)\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x+2}\) = 0

c) \(\dfrac{4}{x-2}\)-x+2 = 0

Answer 1

a.

\(\dfrac{4x-17}{2x^2+1}=0\)

\(\Leftrightarrow4x-17=0\)

\(\Leftrightarrow4x=17\)

\(\Leftrightarrow x=\dfrac{17}{4}\)

Vậy........

Answer 2

b. ĐKXĐ: x khác -2

\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x+2}=0\)

\(\Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\)

\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)

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

Vậy.......

Answer 3

a.phân tích tử số,ta có :

\(4x-17=0=>4x=17=>x=\dfrac{17}{4}\)