a: =>4x=12
hay x=3
b: \(\Leftrightarrow x^2+x-\left(x^2-x-6\right)=7\)
\(\Leftrightarrow x^2+x-x^2+x+6=7\)
=>2x=1
hay x=1/2
c: \(\Leftrightarrow\left(x-3\right)\left(x-1\right)=x^2\)
=>-4x+3=0
=>-4x=-3
hay x=3/4
d: \(\Leftrightarrow\left(2x+2\right)^2-\left(5x-10\right)^2=0\)
\(\Leftrightarrow\left(2x+2-5x+10\right)\left(2x+2+5x-10\right)=0\)
\(\Leftrightarrow\left(-3x+12\right)\left(7x-8\right)=0\)
=>x=4 hoặc x=8/7
a, \(4x=12\Leftrightarrow x=3\)
b, \(x^2+x-x^2+x+6=7\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
c, đk x khác 1 ; -1
\(\Rightarrow x^2-4x+3=x^2\Leftrightarrow x=\dfrac{3}{4}\)(tm)
d, \(4\left(x+1\right)^2-25\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+2\right)^2-\left(5x-10\right)^2=0\Leftrightarrow\left(-3x+12\right)\left(7x-8\right)=0\Leftrightarrow x=-4;x=\dfrac{8}{7}\)
\(a,4x-12=0\Rightarrow4x=12\Rightarrow x=3\\ x\left(x+1\right)-\left(x+2\right)\left(x+3\right)=x^2+x-\left(x^2+5x+6\right)=-4x+6=7\Rightarrow-4x=1\Leftrightarrow x=\dfrac{-1}{4}\)
a,
\(\Leftrightarrow x=\dfrac{0+12}{4}=3\)
b,
\(\Leftrightarrow x^2+x-\left(x^2-3x+2x-6\right)-7=0\)
\(\Leftrightarrow x^2+x-x^2+3x-2x+6-7=0\)
\(\Leftrightarrow2x-1=0\Rightarrow x=\dfrac{0+1}{2}=\dfrac{1}{2}\)
c,
\(\Leftrightarrow\dfrac{x-3}{x+1}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x^2-x-3x+3-x^2=0\)
\(\Leftrightarrow3-4x=0\Rightarrow x=\dfrac{3-0}{4}=\dfrac{3}{4}\)
