Tìm x , biết :
a, x2-9=2(x+3)2
\(=>x^2-9-2\left(x^2+6x+9\right)=0\)
\(=>x^2-9-2x^2-12x-18=0\)
\(=>-x^2-12x-27=0\)
\(=>-\left(x^2+12x+27\right)=0\)
\(=>-\left(x^2+3x+9x+27\right)=0\)
\(=>\left(x+9\right)\left(x+3\right)=0\)
\(=>x+9=0\) hay \(x+3=0\)
\(=>x=-9\) hay \(x=-3\)
Vậy x=-9 hay x = - 3
b, 4x2-4x+1=(5-x)2
\(=>\left(2x-1\right)^2-\left(5-x\right)^2=0\)
\(=>\left(2x-1-5+x\right)\left(2x-1+5-x\right)=0\)
\(=>4\left(3x-6\right)=0\)
\(=>3x-6=0=>3x=6=>x=2\)
Vậy x = 2
c, 4x2-8x+4=2(1-x)(1+x)
\(=>4x^2-8x+4-2\left(1-x^2\right)=0\)
\(=>4x^2-8x+4-2-2x^2=0\)
\(=>6x^2-8x+2=0\)
\(=>6x^2-2x-6x+2=0\)
\(=>2x\left(3x-1\right)-2\left(3x-1\right)=0\)
\(=>\left(3x-1\right)\left(2x-2\right)=0\)
\(=>\left(3x-1\right)2\left(x-1\right)=0\)
\(=>\left(3x-1\right)\left(x-1\right)=0\)
\(=>3x-1=0\) hay \(x-1=0\)
\(=>x=\frac{1}{3}\) hay \(x=1\)
Vậy \(x=\frac{1}{3}\) hay \(x=1\)