Lời giải:

$\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}=\sqrt{(\sqrt{10}-1)^2}$

$=|\sqrt{10}-1|=\sqrt{10}-1$

P/s: Lần sau bạn lưu ý đăng đầy đủ yêu cầu đề bài.

\(\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}\\ =\sqrt{\sqrt{10}^2-2.\sqrt{10}.1+1^2}\\ =\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\left|\sqrt{10}-1\right|=\sqrt{10}-1\)

`a)`\(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(ĐK:x\ge1\)

\(\Leftrightarrow3x+1+x-1-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

\(\Leftrightarrow4x-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

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

\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=4x^2-8x+4\)

\(\Leftrightarrow3x^2-3x+x-1=4x^2-8x+4\)

\(\Leftrightarrow x^2-6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) `(tm)`

Vậy \(S=\left\{5;1\right\}\)

`b)`\(\dfrac{\sqrt{x+27}+\sqrt{27-x}}{\sqrt{27+x}-\sqrt{27-x}}=\dfrac{27}{x}\)

\(ĐK:-27\le x\le27;x\ne0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}+\sqrt{27-x}\right)\left(\sqrt{x+27}-\sqrt{27-x}\right)}{\left(\sqrt{27+x}-\sqrt{27-x}\right)\left(\sqrt{27+x}-\sqrt{27-x}\right)}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}\right)^2-\left(\sqrt{27-x}\right)^2}{\left(\sqrt{27+x}-\sqrt{27-x}\right)^2}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{2x}{54-2\sqrt{\left(27+x\right)\left(27-x\right)}}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{x}{27-\sqrt{27^2-x^2}}=\dfrac{27}{x}\)
\(\Leftrightarrow x^2=27^2-\sqrt{27^2-x^2}\)

\(\Leftrightarrow27^2-x^2-\sqrt{27^2-x^2}=0\)

\(\Leftrightarrow\sqrt{27^2-x^2}\left(\sqrt{27^2-x^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}27^2-x^2=0\\\sqrt{27^2-x^2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm27\\x=\pm\sqrt{27^2-1}\end{matrix}\right.\)

Thế vào pt ta được \(x=\pm27\) là thỏa mãn

Vậy \(S=\left\{\pm27\right\}\)

a) ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{3x+1}-4+2-\sqrt{x-1}=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{4-x+1}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{5-x}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}\right)=0\)

+ Nếu \(x-5=0\Leftrightarrow x=5\left(tmđkxđ\right)\)

+ Nếu \(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}=0\Leftrightarrow\dfrac{3}{\sqrt{3x+1}+4}=\dfrac{1}{2+\sqrt{x-1}}\)

\(\Leftrightarrow3\sqrt{x-1}+2=\sqrt{3x+1}\Rightarrow6x-6+12\sqrt{x-1}=0\)

\(\sqrt{x-1}\left(6\sqrt{x-1}+12\right)=0\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\left(tmđkxđ\right)\)

Vậy ....

ĐKXĐ: \(a>0;a\ne1\)

Ta có: \(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\dfrac{a^2+a\sqrt{a}}{\sqrt{a}+1}\right)\)

\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\left(\dfrac{\sqrt{a}+1}{a\sqrt{a}\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\dfrac{4a\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\left(\dfrac{1}{a\sqrt{a}}\right)\)

\(=\dfrac{4}{a-1}\)

Để \(P>2\) thì \(\dfrac{4}{a-1}>2\Leftrightarrow\dfrac{4}{a-1}-2>2\)

\(\Leftrightarrow\dfrac{2\left(a-3\right)}{1-a}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-3>0\\1-a>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-3< 0\\1-a< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>3\\a< 1\end{matrix}\right.\\\left\{{}\begin{matrix}a< 3\\a>1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow1< a< 3\)

Vậy khi \(1< a< 3\) thì \(P>2\)

\(R=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x>=0;x\ne1\right)\\ =\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x+4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(R=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\dfrac{2}{\sqrt{x}-1}\\ R\in Z=>\dfrac{2}{\sqrt{x}-1}\in Z\\ =>2⋮\left(\sqrt{x}-1\right)=>\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ =>\sqrt{x}\in\left\{2;3;0;-1\right\}\\ =>x\in\left\{4;9;0\right\}\left(TMDK\right)\)

\(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\left(a>0,b>0,a\ne b\right)\\ =\dfrac{\sqrt{ab}.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\\ =\left(\sqrt{a}+\sqrt{b}\right).\left(\sqrt{a}-\sqrt{b}\right)=a-b\)

\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right).\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(a>=0;a\ne1\right)\\ =\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right).\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)

Để đồ thị hai hàm số là các đường thẳng song song : 

$\{\begin{array}{c}m+1=2\\ -m^2-m\ne -2\end{array}$

$\iff \{\begin{array}{c}m=1\\ -m^2-m+2\ne 0\end{array}$

$\iff \{\begin{array}{c}m=1\left(l\right)\\ m\ne 1\\ m\ne -2\end{array}$

Không tồn tại giá trị của m để hai hàm số..........

b/ Hai đường thẳng trên cắt nhau tại 1 điểm trên trục tung khi

\(\left(m+2\right)x+2m^2+1=3x+3\) với x=0 và \(m+2\ne3\Rightarrow m\ne1\)

\(\Rightarrow2m^2=2\Rightarrow m^2=1\Rightarrow m=-1\)

ĐKXĐ: \(a>0;a\ne4\)

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

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

\(=\sqrt{a}.\left(\sqrt{a}-1\right)\)

Vậy \(P=\sqrt{a}\left(\sqrt{a}-1\right)\)

2/3+(-9/16)

=2/3-9/16

=  (32-27)/48

=5/48