a, \(M=2x^3+xy^2-3xy+1\)

b, Thay x = -1 ; y = 2 ta được 

M = -2 - 2 + 6 + 1 = 3 

a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì

\(M=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}-2}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)

\(=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\left(\sqrt{2}-1\right)^2=3-2\sqrt{2}\)

c: M>0

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\)

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-2>0\)

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

=>x>4

a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

 \(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)

c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)

A = 3x^3 +6x^2 + 3xy^3

x= 1 phần 2 ;  p = -1 phần 3

A=3.1 phần 2^3 . -1 phần 3     + 6.(1 phần 2)^2 . (-1 Phần 3)^2+3 1 phần 2 . (-1 phần 3)^3

=-1 phần 8      + -1 phần 2 - 1 phần 2

= -1 phần 4

`a)`

`M=2xy+9xy^2-2xy-7xy^2-3`

`M=(2xy-2xy)+(9xy^2-7xy^2)-3`

`M=2xy^2-3`

___________________________________

`b)` Thay `x=-1;y=2` vào `M`. Ta có:

 `M=2.(-1).2^2-3`

 `M=-2.4-3=-8-3=-11`

a: \(M=\dfrac{x+4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)

c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9

`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)

\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)

\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)

\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

`b,`Ta có :

 \(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)

\(M=\dfrac{3\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(3+\sqrt{2}\right)^2}-3}\)

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

\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)

\(M=\dfrac{6+9\sqrt{2}}{2}\)

`c,`  Để `M<1` Ta có :

 \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)

\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)

\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)

\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )

\(\sqrt{x}< 3\)

\(x< 9\)

Đối chiếu ĐKXĐ ta có : `0<x<9`

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

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

Khi \(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)