\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

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

           \(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)

Bài 1 : 

\(A=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\\ =\sqrt{3}+2+2-\sqrt{3}=4\)

Bài 2 : 

a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)

b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)

Bài 3 : 

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

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{5\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+5\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\dfrac{3\sqrt{x}+1}{3}\)

\(=\dfrac{3x}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)

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

b) Ta có: \(9x^2-10x+1=0\)

\(\Leftrightarrow\left(9x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào P, ta được:

\(P=\dfrac{1}{3-1}=\dfrac{1}{2}\)

c) Thay \(x=8-2\sqrt{7}\) vào P, ta được:

\(P=\dfrac{8-2\sqrt{7}}{3\left(\sqrt{7}-1\right)-1}=\dfrac{8-2\sqrt{7}}{3\sqrt{7}-4}\)

\(=\dfrac{-10+16\sqrt{7}}{47}\)

a)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-4\right)+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(P=\dfrac{3x-2\sqrt{x}-1-3\sqrt{x}+4+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(P=\dfrac{3\left(x+1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

\(P=\dfrac{x+1}{3\sqrt{x}-1}\)

b) Từ phương trình suy ra \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)

Vói x=1 

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

Với x= 1/9

\(P=\dfrac{\dfrac{1}{9}}{3\sqrt{\dfrac{1}{9}}-1}\) không có nghiệm

\(A=\frac{\sqrt{x^2-10x+25}}{x-5}=\frac{\sqrt{\left(x-5\right)^2}}{x-5}\left(ĐK:x\ne5\right)\)

\(A=\frac{\left|x-5\right|}{x-5}\Rightarrow A=\hept{\begin{cases}\frac{x-5}{x-5}=1\\\frac{5-x}{x-5}=-1\end{cases}}\)

\(B=\frac{\sqrt{x=7+6\sqrt{x-2}}}{3+x\sqrt{2}}\)

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

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

1) Ta có: \(\dfrac{a-6\sqrt{a}+9}{5\sqrt{a}-15}\)

\(=\dfrac{\left(\sqrt{a}-3\right)^2}{5\left(\sqrt{a}-3\right)}\)

\(=\dfrac{\sqrt{a}-3}{5}\)

2) Ta có: \(5x-\sqrt{x^2-10x+25}\)

\(=5x-\left|x-5\right|\)

\(=5x-5+x\)

=6x-5

3) Ta có: \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\)

\(=\dfrac{\left|x-1\right|}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{\pm1}{x+1}\)

4) Ta có: \(3\sqrt{5}-\sqrt{46-6\sqrt{5}}\)

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

=1

a: \(P=\dfrac{x-13}{\sqrt{x-9}-2}\)

\(=\dfrac{x-9-4}{\sqrt{x-9}-2}\)

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

b: \(\sqrt{x-9}>=0\forall x\) thỏa mãn ĐKXĐ

=>\(\sqrt{x-9}+2>=2\forall x\) thỏa mãn ĐKXĐ

=>P>=2 với mọi x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-9=0

=>x=9

5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)

\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}-2}{\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\) (ĐK: \(x\ne1;x\ge0\))

\(A=\dfrac{15\sqrt{x}-11}{x+3\sqrt{x}-\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\dfrac{15\sqrt{x}-11}{\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{\left(15\sqrt{x}-11\right)-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(A=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(A=\dfrac{-\left(5\sqrt{x}-2\right)}{\sqrt{x}+3}\)

\(A=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

A = \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)