Bài 1:
a: \(9x^2\cdot\left(2x-3\right)=0\)
=>\(x^2\left(2x-3\right)=0\)
=>\(\left[{}\begin{matrix}x^2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: (x-1)(3x-6)=0
=>3(x-1)(x-2)=0
=>(x-1)(x-2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
c: (x+2)(3-3x)=0
=>3(1-x)(x+2)=0
=>(1-x)(x+2)=0
=>\(\left[{}\begin{matrix}1-x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\end{matrix}\right.\)
d: \(\left(\dfrac{2}{3}x+6\right)\left(8-2x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}x+6=0\\8-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-6\\2x=8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-6:\dfrac{2}{3}=-9\\x=\dfrac{8}{2}=4\end{matrix}\right.\)
e: \(\left(4x+2\right)\left(x^2+1\right)=0\)
=>4x+2=0
=>4x=-2
=>\(x=-\dfrac{2}{4}=-\dfrac{1}{2}\)
f: \(\left(3x-4\right)\left(x+1\right)\left(2x-1\right)=0\)
=>\(\left[{}\begin{matrix}3x-4=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
g: \(\left(3x-2\right)^2\cdot\left(x+1\right)\cdot\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}3x-2=0\\x+1=0\\x-2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-1\\x=2\end{matrix}\right.\)
h: \(\left(2x+3\right)^2=\left(x-5\right)^2\)
=>\(\left(2x+3\right)^2-\left(x-5\right)^2=0\)
=>(2x+3+x-5)(2x+3-x+5)=0
=>(3x-2)(x+8)=0
=>\(\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
i: \(\left(6x-7\right)\left(3x+4\right)=\left(7-6x\right)\left(x-1\right)\)
=>\(\left(6x-7\right)\left(3x+4\right)+\left(6x-7\right)\left(x-1\right)=0\)
=>(6x-7)(3x+4+x-1)=0
=>(6x-7)(4x+3)=0
=>\(\left[{}\begin{matrix}6x-7=0\\4x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
j: \(\left(3x-2\right)\left(x+1\right)=x^2-1\)
=>(3x-2)(x+1)-(x-1)(x+1)=0
=>(x+1)(3x-2-x+1)=0
=>(x+1)(2x-1)=0
=>\(\left[{}\begin{matrix}x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
k: \(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\)
=>\(2\left(4x-1\right)^2+5\left(4x-1\right)\left(x-2\right)=0\)
=>\(\left(4x-1\right)\left[2\left(4x-1\right)+5\left(x-2\right)\right]=0\)
=>\(\left(4x-1\right)\left(8x-2+5x-10\right)=0\)
=>(4x-1)(13x-12)=0
=>\(\left[{}\begin{matrix}4x-1=0\\13x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)
l: \(x^2-8x+12=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
Bài 2:
a: ĐKXĐ: x<>0
\(\dfrac{1}{x}+\dfrac{1}{2x}=\dfrac{3}{2}\)
=>\(\dfrac{2}{2x}+\dfrac{1}{2x}=\dfrac{3}{2}\)
=>\(\dfrac{3}{2x}=\dfrac{3}{2}\)
=>2x=2
=>x=1(nhận)
b: ĐKXĐ: x<>0
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
=>\(x-\dfrac{6}{x}=x+\dfrac{3}{2}\)
=>\(\dfrac{-6}{x}=\dfrac{3}{2}\)
=>x=-4(nhận)
c: ĐKXĐ: x<>3/4
\(\dfrac{3x}{4x-3}=-2\)
=>-2(4x-3)=3x
=>-8x+6=3x
=>-11x=-6
=>\(x=\dfrac{6}{11}\left(nhận\right)\)
d: ĐKXĐ: x<>0
\(\dfrac{3}{8x}-\dfrac{1}{2x}=\dfrac{1}{x^2}\)
=>\(\dfrac{3}{8x}-\dfrac{4}{8x}=\dfrac{1}{x^2}\)
=>\(-\dfrac{1}{8x}=\dfrac{-8}{8x^2}\)
=>\(-\dfrac{x}{8x^2}=\dfrac{-8}{8x^2}\)
=>-x=-8
=>x=8(nhận)
e: ĐKXĐ: x<>2
\(\dfrac{x}{x-2}=\dfrac{2}{x-2}+7\)
=>\(\dfrac{x}{x-2}-\dfrac{2}{x-2}=7\)
=>-1=7(vô lý)
=>Phương trình vô nghiệm
f: ĐKXĐ: \(x\notin\left\{3;-2\right\}\)
\(\dfrac{2}{x-3}=\dfrac{1}{x+2}\)
=>2(x+2)=x-3
=>2x+4=x-3
=>2x-x=-3-4
=>x=-7(nhận)
g: ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};-7\right\}\)
\(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
=>\(\left(6x+1\right)\left(x+7\right)=\left(3x-2\right)\left(2x-3\right)\)
=>\(6x^2+42x+x+7=6x^2-9x-4x+6\)
=>43x+7=-13x+6
=>56x=-1
=>\(x=-\dfrac{1}{56}\left(nhận\right)\)
h: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{2x+1}{x+1}+\dfrac{2}{x}=\dfrac{2}{x\left(x+1\right)}\)
=>\(\dfrac{x\left(2x+1\right)+2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x\left(x+1\right)}\)
=>\(x\left(2x+1\right)+2\left(x+1\right)=2\)
=>\(2x^2+x+2x+2=2\)
=>\(2x^2+3x=0\)
=>x(2x+3)=0
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
