\(Q=\sqrt{x}+\dfrac{25}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{25}{\sqrt{x}}}=10\)

Dấu = xảy ra khi x=25

\(P=\frac{\left(x+7\right)}{\sqrt{x}+3}=\)\(\frac{\left(x+3\sqrt{x}\right)-3\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}\)\(=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}\)

\(=\left(\sqrt{x}+3\right)+\frac{16}{\sqrt{x}+3}-6\)\(\ge8-6=2\)(AM-GM)

''='' <=> x = 1

P\(=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-3\)

ap dung cosi cho 2 so duong \(\left(\sqrt{x}+3\right)va\frac{16}{\sqrt{x}+3}taduoc\)

\(\sqrt{x}+3+\frac{16}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\frac{16}{\sqrt{x}+3}}\)\(\ge2\sqrt{16}=8\)

\(\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-3\ge8-3=5\)

dau = xay ra <=> \(\left(\sqrt{x}+3\right)=\frac{16}{\sqrt{x}+3}\)

<=>  x=1

a) ĐKXĐ:  \(x>0;x\ne9\)

\(A=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{x-9}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\frac{1}{\sqrt{x}+3}\)

b)  \(A=\frac{1}{5}\) \(\Rightarrow\)\(\frac{1}{\sqrt{x}+3}=\frac{1}{5}\)

\(\Rightarrow\)\(\sqrt{x}+3=5\)

\(\Leftrightarrow\)\(\sqrt{x}=2\)

\(\Leftrightarrow\)\(x=4\)(t/m ĐKXĐ)

Vậy...

đề bài có sai chỗ nào k bn???