Bài làm:

đk: \(x>0;x\ne9;x\ne25\)

Ta có: \(\left(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}+\frac{36}{9-x}\right)\div\left(\frac{\sqrt{x}-5}{3\sqrt{x}-x}\right)\)

\(=\left[\frac{\left(3+\sqrt{x}\right)^2-\left(3-\sqrt{x}\right)^2+36}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\div\left[\frac{\sqrt{x}-5}{\left(3-\sqrt{x}\right)\sqrt{x}}\right]\)

\(=\frac{12\sqrt{x}+36}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\frac{\left(3-\sqrt{x}\right)\sqrt{x}}{\sqrt{x}-5}\)

\(=\frac{12\left(3+\sqrt{x}\right)}{3+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{12\sqrt{x}}{\sqrt{x}-5}\)

a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)

b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)

\(\Leftrightarrow\sqrt{x}+1\ge1\)

\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)

\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)

\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi x=0

Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0

ĐKXĐ: ....

\(A=\left(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\right)\)

\(=\left(\frac{3x-6\sqrt{x}-x-2\sqrt{x}+8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{2\sqrt{x}+4-2\sqrt{x}-3}{\sqrt{x}+2}\right)\)

\(=\frac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}+2\right)}{1}=\frac{2x}{\sqrt{x}-2}\)

b/ \(A=\frac{2x}{\sqrt{x}-2}=2\sqrt{x}+4+\frac{8}{\sqrt{x}-2}=2\left(\sqrt{x}-2\right)+\frac{8}{\sqrt{x}-2}+8\ge2\sqrt{\frac{16\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+8=16\)

\(\Rightarrow A_{min}=16\) khi \(\left(\sqrt{x}-2\right)^2=4\Rightarrow x=16\)

\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)

\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{2x}{\sqrt{x}-2}\)

\(\left(\frac{x+3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(\frac{\sqrt{x}\cdot\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\frac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)^2}\)

\(=\frac{-3\cdot\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\)

\(=\frac{-3}{-\left(\sqrt{x}-2\right)}=\frac{3}{\sqrt{x}-2}\)

Chúc bạn học giỏi 

Kết bạn với mình nha

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

1/ \(C=\frac{x+9}{10\sqrt{x}}=\frac{\sqrt{x}}{10}+\frac{9}{10\sqrt{x}}\ge2.\frac{3}{10}=0,6\)

Đạt được khi x = 9

2/ \(E=\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=x-3\sqrt{x}+2\)

\(=\left(x-\frac{2.\sqrt{x}.3}{2}+\frac{9}{4}\right)-\frac{1}{4}\)

\(=\left(\sqrt{x}-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy GTNN là \(-\frac{1}{4}\)đạt được khi \(x=\frac{9}{4}\)

Không có GTLN nhé

3/ Điều kiện xác định bạn tự làm nhé

\(\frac{16}{\sqrt{x}+3}=\frac{-8\sqrt{x}+5}{3\sqrt{x}+1}\)

\(\Leftrightarrow8x+67\sqrt{x}+1=0\)

Tới đây thì bạn xem như phương trình bậc 2 là giải tiếp được. Nhớ đối chiếu điều kiện để loại nghiệm