\(a,x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
=.= hk tốt!!
\(b,4x^2-4x+1=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(5-x\right)^2=0\)
\(\Leftrightarrow\left(2x-1+5-x\right)\left(2x-1-5+x\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
=.= hk tốt!!
\(c,4x^2-8x+4=2\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2-8x+4-2\left(1-x^2\right)=0\)
\(\Leftrightarrow4x^2-8x+4-2+2x^2=0\)
\(\Leftrightarrow6x^2-8x+2=0\)
\(\Leftrightarrow3x^2-4x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)
=.= hk tốt!!
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 \(x=1\)