Question

1/Cho Q=\(\frac{6-a-\sqrt{a}}{\sqrt{a}+3}\)với a≥0

a) Rút gọn

b) Tìm giá trị của a để Q có GTLN

2/Cho a,b,c>0. Rút gọn biểu thức

N=\(\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}\)

Answer 1

1/ \(Q=\frac{\left(2-\sqrt{a}\right)\left(\sqrt{a}+3\right)}{\sqrt{a}+3}=2-\sqrt{a}\)

Do \(\sqrt{a}\ge0\Rightarrow2-\sqrt{a}\le2\Rightarrow Q_{max}=2\) khi \(a=0\)

2/

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

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

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

TH1: Nếu \(a+b\ge c\Rightarrow\sqrt{a+b}-\sqrt{c}\ge0\)

\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{a+b}-\sqrt{c}=2\sqrt{a+b}\)

TH2: Nếu \(a+b< c\Rightarrow\sqrt{a+b}-\sqrt{c}< 0\)

\(\Rightarrow Q=\sqrt{a+b}+\sqrt{c}+\sqrt{c}-\sqrt{a+b}=2\sqrt{c}\)