1.
a)
\(\Leftrightarrow\left[{}\begin{matrix}9x-4=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{9}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
b)
\(\Leftrightarrow\left[{}\begin{matrix}13x+0,26=0\\0,2x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{0,26}{13}=-0,02\\x=\dfrac{4}{0,2}=20\end{matrix}\right.\)
c)
\(\Leftrightarrow\left(x+3\right)\left(2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x+2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+2x-1\right)=0\\ \Leftrightarrow\left(x+2\right)\left(3x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{3}=1\end{matrix}\right.\)
2.
a)
ĐK: \(\left\{{}\begin{matrix}x\ne0\\x\ne-2\end{matrix}\right.\)
Khi đó, pt tương đương: \(3x+6-5x=0\Leftrightarrow-2x+6=0\Leftrightarrow x=-\dfrac{6}{-2}=3\left(tm\right)\)
b)
ĐK: \(\left\{{}\begin{matrix}x\ne\dfrac{1}{2}\\x\ne-\dfrac{5}{2}\end{matrix}\right.\)
Khi đó, pt tương đương: \(2x^2+5x-2x^2+5x-2=0\Leftrightarrow10x-2=0\Leftrightarrow x=\dfrac{2}{10}=\dfrac{1}{5}\left(tm\right)\)
c)
ĐK: \(x\ne2\)
Khi đó, pt tương đương:
\(\dfrac{5x}{x-2}=\dfrac{7\left(x-2\right)}{x-2}+\dfrac{10}{x-2}\\ \Leftrightarrow5x=7x-14+10\Leftrightarrow5x-7x+4=0\\ \Leftrightarrow-2x+4=0\\ \Leftrightarrow x=-\dfrac{4}{-2}=2\left(ktm\right)\)
=> PT vô nghiệm
d)
ĐK: \(x\ne0\)
Khi đó, pt tương đương:
\(\dfrac{2x^2-12}{2x}=\dfrac{2x^2}{2x}+\dfrac{3x}{2x}\\ \Leftrightarrow2x^2-12-2x^2-3x=0\\ \Leftrightarrow-3x-12=0\\ \Leftrightarrow x=\dfrac{12}{-3}=-4\left(tm\right)\)
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;-2\right\}\)
\(\dfrac{1}{x}=\dfrac{5}{3\left(x+2\right)}\)
=>\(5x=3\left(x+2\right)\)
=>5x=3x+6
=>2x=6
=>x=3(nhận)
b: ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-\dfrac{5}{2}\right\}\)
\(\dfrac{x}{2x-1}=\dfrac{x-2}{2x+5}\)
=>x(2x+5)=(2x-1)(x-2)
=>\(2x^2+5x=2x^2-5x+2\)
=>5x=-5x+2
=>10x=2
=>\(x=\dfrac{1}{5}\left(nhận\right)\)
c: ĐKXĐ: x<>2
\(\dfrac{5x}{x-2}=7+\dfrac{10}{x-2}\)
=>\(\dfrac{5x}{x-2}=\dfrac{7x-14+10}{x-2}=\dfrac{7x-4}{x-2}\)
=>7x-4=5x
=>2x=4
=>x=2(loại)
d: ĐKXĐ: x<>0
\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)
=>\(\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)
=>\(2\left(x^2-6\right)=x\left(2x+3\right)\)
=>\(2x^2+3x=2x^2-12\)
=>3x=-12
=>x=-4(nhận)
Bài 1:
a: (9x-4)(2x+5)=0
=>\(\left[{}\begin{matrix}9x-4=0\\2x+5=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{4}{9}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
b: (1,3x+0,26)(0,2x-4)=0
=>1,3(x+0,2)*0,2*(x-20)=0
=>(x+0,2)(x-20)=0
=>\(\left[{}\begin{matrix}x+0,2=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-0,2\\x=20\end{matrix}\right.\)
c: \(2x\left(x+3\right)-5\left(x+3\right)=0\)
=>(x+3)(2x-5)=0
=>\(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
d: \(x^2-4+\left(x+2\right)\left(2x-1\right)=0\)
=>(x-2)(x+2)+(x+2)(2x-1)=0
=>(x+2)(x-2+2x-1)=0
=>(x+2)(3x-3)=0
=>3(x-1)(x+2)=0
=>(x-1)(x+2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
cảm ơn 
