Ta có:
A=\(\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)
=\(\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+3\right)^2}\)
=\(\left|x-3\right|+\left|x+3\right|\)
Xét các TH:
TH 1: \(x\ge3\) ta có: \(\left|x-3\right|=x-3,\left|x+3\right|=x+3\)
A=(x-3)+(x+3)=2x
TH 2:\(-3\le x\le3\) ta có: \(\left|x-3\right|=3-x,\left|x+3\right|=x+3\)
A=(3-x)+(x+3)=6
TH 3: \(x\le-3\) ta có: \(\left|x-3\right|=3-x,\left|x+3\right|=-x-3\)
A=(3-x)+(-x-3)=-2x
(A=sqrt{x^2-6x+9}+sqrt{x^2+6x+9})
(A=sqrt{x^2-3x-3x+9}+sqrt{x^2+3x+3x+9})
(A=sqrt{left(x-3 ight)^2}+sqrt{left(x+3 ight)^2})
(A=x-3+x+3=2x)
\(A=\sqrt{x^2-6x+9}+\sqrt{x^2+6x+9}\)
\(A=\sqrt{x^2-3x-3x+9}+\sqrt{x^2+3x+3x+9}\)
\(A=\sqrt{\left(x^2-3x\right)-\left(3x-9\right)}+\sqrt{\left(x^2+3x\right)+\left(3x+9\right)}\)
\(A=\sqrt{x.\left(x-3\right)-3.\left(x-3\right)}+\sqrt{x.\left(x+3\right)+3.\left(x+3\right)}\)
\(A=\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+3\right)^2}\)
\(A=\left(x-3\right)+\left(x+3\right)=x-3+x+3=2x\)
Chúc bạn học tốt!!!
Có phải thầy phynit là người tick đúng cho bạn Q.Duy không ạ?
