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Ta có:

Vậy 

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

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

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

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

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

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 : \(M=\frac{2m}{9}\) 

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

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

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

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

\(=\frac{\sqrt{m}}{\left|1-x\right|}.\frac{2\sqrt{m}.\left|1-x\right|}{9}=\frac{2m}{9}\)

a: =2-81-24-81=-22-162=-184

a) Bạn Nguyễn Lê Phước Thịnh giải rồi nha bạn :)

b) (-9-x+2)+9

=-9-x+2+9

=-9+(-x)+2+9

=(-x)+(-9+2+9)

=(-x)+2

c) 66-(12-x)+(12-66)

=66-12+x+12-66

=x

d) 15-(15-x+93)+93

=15-15+x-93+93

=x

\(A\left(x\right)=\dfrac{4x^4+81}{2x^2-6x+9}\)

\(=\dfrac{4x^4+36x^2+81-36x^2}{2x^2-6x+9}\)

\(=\dfrac{\left(2x^2+9\right)^2-\left(6x\right)^2}{2x^2+9-6x}\)

\(=\dfrac{\left(2x^2+9+6x\right)\left(2x^2+9-6x\right)}{2x^2+9-6x}\)

\(=2x^2+6x+9\)

=>\(M\left(x\right)=2x^2+6x+9\)

\(=2\left(x^2+3x+\dfrac{9}{2}\right)\)

\(=2\left(x^2+3x+\dfrac{9}{4}+\dfrac{9}{4}\right)\)

\(=2\left(x+\dfrac{3}{2}\right)^2+\dfrac{9}{2}>=\dfrac{9}{2}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{2}=0\)

=>\(x=-\dfrac{3}{2}\)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

1:

a: \(\sqrt{25}+\sqrt{49}=5+7=12\)

b: \(\sqrt{121}-\sqrt{81}=11-9=2\)

2: x>-2

=>2x>-4

=>2x+1>-3

=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa

3:

a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)

\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)

\(=\sqrt{3}-1-\sqrt{3}=-1\)

b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}=21\)

c:

\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)