Question

Giải phương trình:

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

mk giải mãi ko ra

Answer 1

Ta có :

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

<=> \(\frac{x^2+6}{x}\) = \(\frac{2x+3}{2}\)

<=> \(\frac{2\left(x^2+6\right)}{2x}\)= \(\frac{x\left(2x+3\right)}{2x}\)

<=> 2x² + 12 = 2x² + 3x

<=> 2x² + 12 - 2x² - 3x = 0

<=> 12 - 3x = 0

<=> -3x = -12

<=> x = \(\frac{-12}{-3}\) = 4

Vậy x = 4 thì \(\frac{x^2+6}{x}\) = x + \(\frac{3}{2}\)

Answer 2

Ta có:

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

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

\(\Rightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-12=2x^2+3x\)

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

\(\Leftrightarrow-12-3x=0\)

\(\Leftrightarrow-3x=12\)

\(\Leftrightarrow x=\frac{12}{-3}\)

\(\Leftrightarrow x=-4\)

Vậy với \(x=-4\) thì \(\frac{x^2-6}{x}=x+\frac{3}{2}\)

Answer 3

Quên, điều kiện xác định là \(x\ne0\)

Answer 4

Ta có :

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

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

<=> \(\frac{2\left(x^2-6\right)}{2x}\) = \(\frac{x\left(2x+3\right)}{2x}\)

<=> 2x² - 12 = 2x² + 3x

<=> 2x² - 12 - 2x² - 3x = 0

<=> - 12 - 3x =0

<=> -3x = 12

<=> x = \(\frac{12}{-3}\) = -4

Vậy x = -4 thì \(\frac{x^2-6}{x}\) = x + \(\frac{3}{2}\)