Question

[Nguyễn Hi]

Giúp mình với ạ

Sử dụng Cosi để giải nhé

Tìm GTNN của B

B=x+5/căn x+2 (với x>0)

Answer 1

[Phạm Thị Thùy Linh]

\(đkcđ\Leftrightarrow x\ge0\)

\(B=\frac{x+5}{\sqrt{x}+2}=\frac{x-4+9}{\sqrt{x}+2}=\frac{x-4}{\sqrt{x}+2}+\frac{9}{\sqrt{x}+2}.\)

\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{9}{\sqrt{x}+2}=\sqrt{x}-2+\frac{9}{\sqrt{x}+2}\)

\(=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\)

Áp dụng bđt Cô - si cho hai số dương \(\sqrt{x}+2\)và \(\frac{9}{\sqrt{x}+2}\), ta có :

\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge2\sqrt{\frac{\left(\sqrt{x}+2\right).9}{\sqrt{x}+2}}\)

\(\Rightarrow\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge2.3\)

\(\Rightarrow\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\ge6-4\)

\(\Rightarrow\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\ge2\)

Hay \(B_{min}=2\)\(\Leftrightarrow\sqrt{x}+2=\frac{9}{\sqrt{x}+2}\)

\(\Rightarrow\sqrt{x}+2-\frac{9}{\sqrt{x}+2}=0\)

\(\Rightarrow\frac{\left(\sqrt{x}+2\right)^2-9}{\sqrt{x}+2}=0\)

\(\Rightarrow\left(\sqrt{x}+2\right)^2-3^2=0\)

\(\Rightarrow\left(\sqrt{x}+2-3\right)\left(\sqrt{x}+2+3\right)=0\)

\(\Rightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)=0\)

Vì \(\sqrt{x}+5>0\Rightarrow\sqrt{x}-1=0\)

\(\Rightarrow\sqrt{x}=1\Rightarrow x=1\)

\(KL:B_{min}=2\Leftrightarrow x=1\)