a) Ta có: \(9x^2-3=\left(3x+1\right)\left(2x-3\right)\)
⇔\(9x^2-3=6x^2-9x+2x-3\)
\(\Leftrightarrow9x^2-3=6x^2-7x-3\)
\(\Leftrightarrow9x^2-3-6x^2+7x+3=0\)
\(\Leftrightarrow3x^2+7x=0\)
\(\Leftrightarrow x\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-7}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-7}{3}\right\}\)
b) ĐKXĐ: x∉{0;5}
Ta có: \(\frac{3x}{x-5}+\frac{1}{x}=\frac{4x+3}{x\left(x-5\right)}+3\)
\(\Leftrightarrow\frac{3x^2}{x\left(x-5\right)}+\frac{x-5}{x\left(x-5\right)}=\frac{4x+3}{x\left(x-5\right)}+\frac{3x\left(x-5\right)}{x\left(x-5\right)}\)
⇔\(3x^2+x-5=4x+3+3x\left(x-5\right)\)
\(\Leftrightarrow3x^2+x-5=4x+3+3x^2-15x\)
\(\Leftrightarrow3x^2+x-5-3x^2+11x-3=0\)
\(\Leftrightarrow12x-8=0\)
\(\Leftrightarrow12x=8\)
hay \(x=\frac{2}{3}\)(tm)
Vậy: \(x=\frac{2}{3}\)
Bài làm
a) 9x2 - 3 = ( 3x + 1 )( 2x - 3 )
<=> 9x2 - 3 = 6x2 - 9x + 2x - 3
<=> 9x2 - 6x2 + 9x - 2x = 0
<=> 3x2 + 7x = 0
<=> x( 3x + 7 ) = 0
<=> \(\left[{}\begin{matrix}x=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{7}{3}\end{matrix}\right.\)
Vậy S = { 0; -7/3 }
b) \(\frac{3x}{x-5}+\frac{1}{x}=\frac{4x+3}{x\left(x-5\right)}+3\) ĐKXĐ: \(x\ne0;x\ne5\)
\(\Leftrightarrow\frac{3x^2}{x\left(x-5\right)}+\frac{x-5}{x\left(x-5\right)}=\frac{4x+3}{x\left(x-5\right)}+\frac{3x\left(x-5\right)}{x\left(x-5\right)}\)
\(\Rightarrow3x^2+x-5=4x+3+3x^2-15x\)
\(\Leftrightarrow3x^2+x-5-4x-3-3x^2+15x=0\)
\(\Leftrightarrow12x-8=0\)
\(\Leftrightarrow x=\frac{8}{12}=\frac{2}{3}\)( TM )
Vậy x = 2/3 là nghiệm phương trình.
