Áp dụng bất đẳng thức bunhiacopxki ta có: \(A^2=\left(6\sqrt{x-2}+8\sqrt{5-x}\right)^2\le\left(6^2+8^2\right)\left(x-2+5-x\right)\)

\(A^2\le300\Leftrightarrow-\sqrt{300}\le A\le\sqrt{300}\)

\(\frac{2\sqrt{2}\left(1+\sqrt{3}\right)}{\frac{3\left(1+\sqrt{3}\right)}{\sqrt{2}}}=\frac{2\sqrt{2}\sqrt{2}\left(1+\sqrt{3}\right)}{3\left(1+\sqrt{3}\right)}=\frac{4}{3}\)

\(\frac{2\sqrt{2}\left(1+\sqrt{3}\right)}{3\sqrt{\frac{4+2\sqrt{3}}{2}}}=\frac{2\sqrt{2}\left(1+\sqrt{3}\right)}{3\sqrt{\frac{3+2\sqrt{3}+1}{2}}}=\frac{2\sqrt{2}\left(1+\sqrt{3}\right)}{3\sqrt{\frac{\left(1+3\right)^2}{2}}}\)

Còn lại bạn giải tiếp đc chứ :D 

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

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

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

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

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

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

\(M=3\)

b) \(\sqrt{x}=M\)

\(\Leftrightarrow x=M^2\)

thay vào ta có: 

\(x=3^2\)

\(x=9\)

c) \(M=3\in N\)

\(\Rightarrow x=3\)

d) \(M>1\Leftrightarrow x>1\)

Người lạ xin kết bạn !

Kb vs e nha a , e hok lop 7!😊