a: \(M=A+B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}-1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

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

\(=\dfrac{3x+7\sqrt{x}-6}{x-9}\)

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

b: M=M^4

=>M=0 hoặc M=1

=>3 căn x-2=căn x-3 hoặc 3 căn x-2=0

=>x=4/9

Bài `1`

\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)

2:

a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

b: B=5

=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)

=>\(5\sqrt{x}+15=\sqrt{x}+8\)

=>\(4\sqrt{x}=-7\)(loại)

Vậy: \(x\in\varnothing\)

Do \(a\le\left|a\right|,b\le\left|b\right|\) nên ta chỉ cần chứng minh

\(\dfrac{\left|a\right|}{\sqrt{a^2+b^2}}+\dfrac{\left|b\right|}{\sqrt{9a^2+b^2}}+\dfrac{2\left|a\right|\left|b\right|}{\sqrt{a^2+b^2}.\sqrt{9a^2+b^2}}\le\dfrac{3}{2}\)

Đặt \(a^2=x,b^2=3y^2\)

\(P=2\sqrt{\dfrac{x}{x+3y}}+2\sqrt{\dfrac{y}{y+3x}}+4\sqrt{\dfrac{xy}{\left(x+3y\right)\left(y+3x\right)}}\le3\)

Sử dụng BĐT AM-GM, ta có

\(2\sqrt{\dfrac{x}{x+3y}}\le\dfrac{x}{x+y}+\dfrac{x+y}{3x+y},2\sqrt{\dfrac{y}{y+3x}}\le\dfrac{y}{x+y}+\dfrac{x+y}{y+3x}\)\(4\sqrt{\dfrac{xy}{\left(x+3y\right)\left(y+3x\right)}}\le\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}+\dfrac{1}{2}\)

Cộng ba bất đẳng thức trên vế theo vế

\(P\le\dfrac{3}{2}+\dfrac{x+y}{x+3y}+\dfrac{x+y}{y+3x}+\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}\)

Và do đó chứng minh sẽ hoàn tất nếu ta chỉ ra được rằng:

\(\dfrac{x+y}{x+3y}+\dfrac{x+y}{y+3x}+\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}\le\dfrac{3}{2}\)

Ta có: \(\dfrac{3}{2}-\dfrac{x+y}{x+3y}-\dfrac{x+y}{y+3x}-\dfrac{8xy}{\left(x+3y\right)\left(y+3x\right)}=\dfrac{3}{2}-\dfrac{4\left(x+y\right)^2+8xy}{\left(x+3y\right)\left(y+3x\right)}=\dfrac{\left(x-y\right)^2}{2\left(x+3y\right)\left(y+3x\right)}\ge0\)Bài toán được chứng minh xong. Đẳng thức xảy ra khi \(b=\sqrt{3}a>0\)

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

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

Lời giải:

a.

\(B=\frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-2x}{(\sqrt{x}+3)(\sqrt{x}-3)}=\frac{x-3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)

b.

\(P=AB=\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)

Để $P<0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+3}<0$

Mà $\sqrt{x}+3>0$ nên $\sqrt{x}-2<0$

$\Leftrightarrow 0< x< 4$

Kết hợp với ĐKXĐ suy ra $0< x< 4$

Mà $x$ nguyên nên $x\in left\{1; 2; 3\right\}$