Question

Answer 1

\(\sqrt{x^2-6x+9}+\sqrt{2x^2+8x+8}=\sqrt{x^2-2x+1}\)

=>\(\sqrt{\left(x-3\right)^2}+\sqrt{2\left(x+2\right)^2}=\sqrt{\left(x-1\right)^2}\)

=>|x-3|+\(\sqrt2\left|x+2\right|=\left|x-1\right|\)

=>\(\sqrt2\left|x+2\right|+\left|x-3\right|-\left|x-1\right|=0\left(1\right)\)

TH1: x<-2

(1) sẽ trở thành: \(\sqrt2\left(-x-2\right)+\left(3-x\right)-\left(1-x\right)=0\)

=>\(-\sqrt2\left(x+2\right)+3-x-1+x=0\)

=>\(-\sqrt2\left(x+2\right)=-2\)

=>\(x+2=\sqrt2\)

=>\(x=\sqrt2-2\) (loại)

TH2: -2<=x<1

(1) sẽ trở thành: \(\sqrt2\left(x+2\right)+\left(3-x\right)-\left(1-x\right)=0\)

=>\(\sqrt2\left(x+2\right)+3-x-1+x=0\)

=>\(\sqrt2\left(x+2\right)=-2\)

=>\(x+2=-\sqrt2\)

=>\(x=-\sqrt2-2\) (loại)

TH3: 1<=x<3

(1) sẽ trở thành: \(\sqrt2\left(x+2\right)+\left(3-x\right)-\left(x-1\right)=0\)

=>\(x\sqrt2+2\sqrt2+3-x-x+1=0\)

=>\(x\left(\sqrt2-2\right)=-4-2\sqrt2\)

=>\(x=\frac{-4-2\sqrt2}{\sqrt2-2}=\frac{4+2\sqrt2}{2-\sqrt2}=\frac{\left(4+2\sqrt2\right)\left(2+\sqrt2\right)}{\left(2-\sqrt2\right)\left(2+\sqrt2\right)}\)

=>\(x=\frac{8+4\sqrt2+4\sqrt2+4}{4-2}=\frac{12+8\sqrt2}{2}=6+4\sqrt2\) (loại)

TH4: x>=3

(1) sẽ trở thành: \(\sqrt2\left(x+2\right)+\left(x-3\right)-\left(x-1\right)=0\)

=>\(\sqrt2\left(x+2\right)+x-3-x+1=0\)

=>\(\sqrt2\left(x+2\right)=2\)

=>\(x+2=\sqrt2\)

=>\(x=\sqrt2-2\) (loại)