Bài 1:
a: ĐKXĐ: \(4-x^2\ne0\)
=>\(x^2\ne4\)
=>\(x\notin\left\{2;-2\right\}\)
b: ĐKXĐ: \(x-3\ne0\)
=>\(x\ne3\)
c: ĐKXĐ: \(\left\{{}\begin{matrix}-5x+5\ne0\\x^2-1\ne0\end{matrix}\right.\)
=>\(x\notin\left\{1;-1\right\}\)
Bài 2:
a: \(\left(3x+5\right)\left(\dfrac{12}{5}-2x\right)=0\)
=>\(\left[{}\begin{matrix}3x+5=0\\\dfrac{12}{5}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-5\\2x=\dfrac{12}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{6}{5}\end{matrix}\right.\)
b: \(\left(7x-1\right)^2=4\left(1-2x\right)^2\)
=>\(\left(7x-1\right)^2=\left(4x-2\right)^2\)
=>\(\left(7x-1-4x+2\right)\left(7x-1+4x-2\right)=0\)
=>(3x+1)(11x-3)=0
=>\(\left[{}\begin{matrix}3x+1=0\\11x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{3}{11}\end{matrix}\right.\)
c:
ĐKXĐ: x<>-3/4
\(\dfrac{2x^2}{4x+3}-\dfrac{4x-3}{8}=1\)
=>\(\dfrac{8\cdot2x^2-\left(4x-3\right)\left(4x+3\right)}{8\left(4x+3\right)}=1\)
=>\(8\left(4x+3\right)=16x^2-16x^2+9=9\)
=>32x+24=9
=>32x=9-24=-15
=>\(x=-\dfrac{15}{32}\left(nhận\right)\)
d: ĐKXĐ: \(x\notin\left\{1;-5\right\}\)
\(\dfrac{x}{x^2+4x-5}-\dfrac{2}{x-1}=0\)
=>\(\dfrac{x}{\left(x+5\right)\left(x-1\right)}-\dfrac{2}{x-1}=0\)
=>\(\dfrac{x-2\left(x+5\right)}{\left(x+5\right)\left(x-1\right)}=0\)
=>x-2(x+5)=0
=>x-2x-5=0
=>-x-5=0
=>x=-5(loại)
