a) ĐKXĐ: \(a>1;a\ne-1\) 

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

\(\Leftrightarrow B=\dfrac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}.\dfrac{\sqrt{1+a}.\sqrt{1-a}}{3+\sqrt{1+a}.\sqrt{1-a}}\)

\(\Leftrightarrow B=\sqrt{1-a}\)

b) Thay a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\) vào B ta được:

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

\(\Leftrightarrow B\) \(=\sqrt{\dfrac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}}\)

\(\Leftrightarrow B\) \(=\sqrt{\dfrac{2}{2+\sqrt{3}}}\) 

\(\Leftrightarrow B\)\(=\sqrt{\dfrac{4}{4+2\sqrt{3}}}\) \(\Leftrightarrow B\) \(=\dfrac{\sqrt{4}}{\sqrt{3+2\sqrt{3}+1}}\) 

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

c) Có \(\sqrt{B}>B\) \(\Leftrightarrow\sqrt{\sqrt{1-a}}>\sqrt{1-a}\) 

\(\Leftrightarrow\sqrt{1-a}>1-a\) 

\(\Leftrightarrow\sqrt{1-a}-\left(1-a\right)>0\) 

\(\Leftrightarrow\sqrt{1-a}.\left(1-\sqrt{1-a}\right)>0\)

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

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

1) ĐKXĐ: \(a\ge0;a\ne1\)

\(P=\left[\dfrac{a+\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)\(:\left[\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)

\(\Leftrightarrow P=\left[\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)+2.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right]\)\(:\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)

\(\Leftrightarrow P=\left[\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right].\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

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

\(\Leftrightarrow P=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

2) Có : \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}-\dfrac{\sqrt{a}+9}{8}\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{a}-\left(\sqrt{a}+9\right).\left(\sqrt{a}+1\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{a}-a-10\sqrt{a}-9}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{-\left(a-6\sqrt{a}+9\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}\le0\)

Vì \(\sqrt{a}\ge0\Rightarrow8.\left(\sqrt{a}+1\right)>0\)  mà \(\left(\sqrt{a}-3\right)^2\) \(\ge0\) 

\(\Rightarrow\) \(\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}=0\) \(\Rightarrow\left(\sqrt{a}-3\right)^2=0\) \(\Leftrightarrow\sqrt{a}-3=0\Leftrightarrow\sqrt{a}=3\Leftrightarrow a=9\)

Vậy để\(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\) thì \(a=9\)

Giải phương trình:

\(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)

chúc bạn học tốt!

10)\(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{\left(3-x\right).\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{\left(x-1\right).\left(3-x\right)}{\left(x+3\right).\left(3-x\right)}+\dfrac{x.\left(x+3\right)}{\left(3-x\right).\left(x+3\right)}=\dfrac{7x-3}{\left(x+3\right).\left(3-x\right)}\)

\(\Leftrightarrow\dfrac{\left(x-1\right).\left(3-x\right)+x.\left(x+3\right)}{\left(x+3\right).\left(3-x\right)}=\dfrac{7x-3}{\left(x+3\right).\left(3-x\right)}\)

\(\Rightarrow3x-x^2-3+x+x^2+3x=7x-3\)

\(\Leftrightarrow7x-3=7x-3\) (đúng với mọi x) .

Vậy phương trình vô số nghiệm.

7) \(\left|2-3x\right|=2x+1\) 

Có \(\left|2-3x\right|=\) \(2-3x\) nếu \(2-3x\ge0\Leftrightarrow x\le\dfrac{2}{3}\) 

      \(\left|2-3x\right|=\) 3x-2 nếu \(2-3x< 0\Leftrightarrow x>\dfrac{2}{3}\) 

Trường hợp 1: nếu \(\left|2-3x\right|=\) \(2-3x\) thì ta được:

\(2-3x=2x+1\Leftrightarrow1=5x\Leftrightarrow x=\dfrac{1}{5}\)

Trường hợp 2 nếu \(\left|2-3x\right|=\) 3x-2 thì ta được:

\(3x-2=2x+1\Leftrightarrow3=x\)

Vậy \(x\in\left\{\dfrac{1}{5};3\right\}\) 

8) \(\left|2x-1\right|+\left|4x^2-1\right|=0\)

\(\Leftrightarrow\left|2x-1\right|+\left|\left(2x-1\right).\left(2x+1\right)\right|=0\)

\(\Leftrightarrow\left|2x-1\right|+\left|2x-1\right|.\left|2x+1\right|=0\)

\(\Leftrightarrow\left|2x-1\right|.\left(1+\left|2x+1\right|\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left|2x-1\right|=0\\1+\left|2x+1\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left|2x+1\right|=-1\end{matrix}\right.\) \(\Rightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)

Vậy x=\(\dfrac{1}{2}\) 

9) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2-x+3x-3}-\dfrac{2x-1}{1-x}\)

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

\(\Leftrightarrow\dfrac{3x+8}{x+3}=\dfrac{4}{\left(x+3\right).\left(x-1\right)}+\dfrac{2x-1}{x-1}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right).\left(x-1\right)}{\left(x+3\right).\left(x-1\right)}=\dfrac{4+\left(2x-1\right).\left(x+3\right)}{\left(x+3\right).\left(x-1\right)}\)

\(\Rightarrow\left(3x+8\right).\left(x-1\right)=4+\left(2x-1\right).\left(x+3\right)\)

\(\Leftrightarrow3x^2-3x+8x-8=4+2x^2+6x-x-3\)

\(\Leftrightarrow3x^2+5x-8=2x^2+5x+1\)

\(\Leftrightarrow x^2-9=0\)

\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

Vậy x\(\in\left\{-3;3\right\}\)

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