\(\frac{2}{3}< \frac{3}{2}\)

Zậy hử 

~Hok tốt~

 2√3 < 3√2

Vì \(\sqrt{x+9}\ge0\)

mà -2<0

Vậy phương trình vô nghiệm

Vì \(\sqrt{x+9\ge0}\)

Mà -2 < 0 

Vậy phương trình vô nghiệm.

vì \(\sqrt{x-9}\ge0\forall x\)

mà \(-2< 0\)

=> phương trình vô nghiệm

chúc bn học tốt

\(\sqrt{x-9}=-2.\)

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

\(\Rightarrow|x-9|=4\)

TH1 : \(x-9=4\Rightarrow x=4+9=13\)

TH2 : \(x-9=-4\Rightarrow x=-4+9=5\)

Vậy \(x\in\left\{5;13\right\}\)

\(1=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\ge\left(a+b\right)\left[\left(a+b\right)^2-\frac{3}{4}\left(a+b\right)^2\right]=\frac{\left(a+b\right)^3}{4}\)

\(\Rightarrow\left(a+b\right)^3\le4\Rightarrow a+b\le\sqrt[3]{4}\)

\(A=\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\le\sqrt{2\sqrt[3]{4}}\)

\("="\Leftrightarrow a=b=\frac{1}{\sqrt[3]{2}}\)

\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3}{2}\sqrt{6}+\frac{2.1}{3}\sqrt{2.3}-\frac{4.1}{2}\sqrt{3.2}\)

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

\(=\sqrt{6}\left(\frac{9}{6}+\frac{4}{6}-\frac{12}{6}\right)=\sqrt{6}.\frac{1}{6}=\frac{\sqrt{6}}{6}\)

Vậy \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}\)

\(M=\sqrt{\frac{m}{1-2x+x^2}}\times\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m\times\left(1-2x+x^2\right)}}{\sqrt{81}}\)

\(=\frac{\sqrt{m}}{\sqrt{1-2x+x^2}}\times\frac{\sqrt{4m}\times\sqrt{1-2x+x^2}}{9}\)

\(=\frac{\sqrt{m}\times\sqrt{4m}}{9}\)

\(=\frac{2m}{9}\)

vậy . . .

\(M=\sqrt{\frac{m}{1-2x+x^2}}.\sqrt{\frac{4m-8mx+4mx^2}{81}}\)

     \(=\sqrt{\frac{m}{\left(1-x\right)^2}}.\sqrt{\frac{4m\left(1-2x+x^2\right)}{81}}\)

     \(=\sqrt{\frac{m}{\left(1-x\right)^2}.\frac{4m\left(1-x\right)^2}{81}}\)

     \(=\frac{\sqrt{4m^2}}{81}\)

     \(=\frac{\sqrt{4m^2}}{\sqrt{81}}=\frac{2m}{9}\)

Vậy : \(M=\frac{2m}{9}\)