a: \(B=1-\sqrt{\left(x-1\right)^2+1}\)

(x-1)^2+1>=1

=>\(\sqrt{\left(x-1\right)^2+1}>=1\)

=>\(B< =0\)

Dấu = xảy ra khi x=1

b: 

ĐKXĐ: -(x+2)^2+2>=0

=>-(x+2)^2>=2

=>(x+2)^2<=2

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

\(-x^2+4x-2=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)=-\left(x-2\right)^2+2< =2\)

=>\(0< =\sqrt{4x-x^2-2}< =\sqrt{2}\)

=>1<=C<=căn 2+1

\(C_{max}=\sqrt{2}+1\Leftrightarrow x=2\)

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

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

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

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

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

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

\(M=3\)

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) \(M>1\Leftrightarrow x>1\)

\(\frac{1}{1-\sqrt{x}}-1\)sao bằng\(\frac{\sqrt{x}}{\sqrt{x}-1}\)đc

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

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

Câu 1 : 

Đk: \(x\ge1\) 

\(\sqrt{x-1}+\sqrt{2x-1}=5\\ \Leftrightarrow x-1+2\sqrt{\left(x-1\right)\left(2x-1\right)}+2x-1=25\\ \Leftrightarrow2\sqrt{2x^2-3x+1}=27-3x\\ \)

\(\Leftrightarrow\begin{cases}27-3x\ge0\\4\left(2x^2-3x+1\right)=9x^2-162x+729\end{cases}\) \(\Leftrightarrow\begin{cases}x\le9\\x^2-150x+725=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x\le9\\x=145hoặcx=5\end{cases}\)

với x= 5 thoản mãn điều kiện, x=145 loại

Vậy \(S=\left\{5\right\}\)

a/ ĐKXĐ: ...

\(\Leftrightarrow\frac{4x+3}{x+1}=9\Leftrightarrow4x+3=9\left(x+1\right)\)

\(\Leftrightarrow5x=-6\Rightarrow x=-\frac{6}{5}\)

b/ ĐKXĐ: \(x\ge0\)

Nhân cả tử và mẫu của từng số hạng với biểu thức liên hợp và rút gọn ra được:

\(\sqrt{x+5}-\sqrt{x+4}+\sqrt{x+4}-\sqrt{x+3}+...+\sqrt{x+1}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+5}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+5}=1+\sqrt{x}\)

\(\Leftrightarrow x+5=x+1+2\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)

c/ \(\Leftrightarrow2xy-6x-5y+15=33\)

\(\Leftrightarrow2x\left(y-3\right)-5\left(y-3\right)=33\)

\(\Leftrightarrow\left(2x-5\right)\left(y-3\right)=33\)

Đến đây là pt ước số đơn giản rồi