Question

\(\sqrt{a+b+9c+6\sqrt{bc+ac}}+\sqrt{a+b+c+9c-6\sqrt{bc+ac}}\)

Dk (ABC)>0

Answer 1

Đặt \(A=\sqrt{a+b+9c+6\sqrt{bc+ac}}+\sqrt{a+b+9c-6\sqrt{bc+ac}}\)

=\(\sqrt{\left(a+b\right)+2.3.\sqrt{c}\sqrt{a+b}+9c}+\sqrt{\left(a+b\right)-2.3.\sqrt{c}.\sqrt{a+b}+9c}\)

=\(\sqrt{\left(\sqrt{a+b}+3\sqrt{c}\right)^2}+\sqrt{\left(\sqrt{a+b}-3\sqrt{c}\right)^2}\)

=\(\sqrt{a+b}+3\sqrt{c}+\left|\sqrt{a+b}-3\sqrt{c}\right|\)(1)

TH1: \(\sqrt{a+b}< 3\sqrt{c}\) hay a+b<9c

Từ (1) <=>A= \(\sqrt{a+b}+3\sqrt{c}+3\sqrt{c}-\sqrt{a+b}=6\sqrt{c}\)

TH2: \(\sqrt{a+b}\ge3\sqrt{c}\) hay \(a+b\ge9c\)

Từ (1) <=> A= \(\sqrt{a+b}+3\sqrt{c}+\sqrt{a+b}-3\sqrt{c}=2\sqrt{a+b}\)

Answer 2

yêu cầu j