\(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x+3}{2x-4}\)
\(\Leftrightarrow\dfrac{x+5}{3\left(x-2\right)}-\dfrac{1}{2}=\dfrac{2x+3}{2\left(x-2\right)}\text{ĐKXĐ:}x\ne2\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x+3\right)}{6\left(x-2\right)}\)
\(\Rightarrow2x+10-3x+6=6x+9\)
\(\Leftrightarrow-x+16=6x+9\)
\(\Leftrightarrow-x-6x=-16+9\)
\(\Leftrightarrow-7x=-7\)
\(\Leftrightarrow x=1\left(\text{nhận}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{1\right\}\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}\text{ĐKXĐ:}x\ne\pm\dfrac{1}{3}\)
\(\Rightarrow12=1-3x-3x+9x^2-1-3x-3x-9x^2\)
\(\Leftrightarrow12=-12x\)
\(\Leftrightarrow x=-1\left(\text{nhận}\right)\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-1\right\}\)
a: =>2(x+5)-3(x-2)=3(2x+3)
=>6x+9=2x+10-3x+6=-x+16
=>7x=7
=>x=1(nhận)
b: =>12=(1-3x)^2-(1+3x)^2
=>1-6x+9x^2-1-6x-9x^2=12
=>-12x=12
=>x=-1(nhận)
