ĐKXĐ: \(a>0;a\ne1\)
\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\dfrac{a-1}{\sqrt{a}+1}\)
\(=\left[\dfrac{\sqrt{a}.\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\dfrac{a-1}{\sqrt{a}+1}\)
\(=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\dfrac{a-1}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-2}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=\sqrt{a}-2\)

1)Để căn có nghĩa \(\Leftrightarrow\dfrac{-a}{3}\ge0\Leftrightarrow a\le0\)

Vậy...

2)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2+1}{1-3a}\ge0\\1-3a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-3a>0\left(vìa^2+1>0\right)\\1-3a\ne0\end{matrix}\right.\)

\(\Leftrightarrow1-3a>0\Leftrightarrow3a< 1\Leftrightarrow a< \dfrac{1}{3}\)

Vậy...

3)Để căn có nghĩa 

\(\Leftrightarrow a^2-6a+10\ge0\Leftrightarrow\left(a^2-6a+9\right)+1\ge0\Leftrightarrow\left(a-3\right)^2+1\ge0\left(lđ;\forall a\right)\)

Vậy căn luôn có nghĩa với mọi a

4)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a-1}{a+2}\ge0\\a+2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+2< 0\end{matrix}\right.\end{matrix}\right.\\a+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a\ge1\\a>-2\end{matrix}\right.\\\left\{{}\begin{matrix}a\le1\\a< -2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -2\end{matrix}\right.\)

Vậy...

\(\dfrac{1+a}{1+\sqrt{a}+1}=\dfrac{\sqrt{6}}{1+\sqrt{6}}\\ \Leftrightarrow\left(1+\sqrt{6}\right)\left(1+a\right)=\sqrt{6}\left(1+\sqrt{a}+a\right)\\ \Leftrightarrow1+a+\sqrt{6}+\sqrt{6}a=\sqrt{6}+\sqrt{6}a+\sqrt{6}a\\ \Leftrightarrow1+a=\sqrt{6}a\\ \Leftrightarrow\sqrt{6}a=1+a\\ \Leftrightarrow6a=1+2a+a^2\\ \Leftrightarrow6a-1-2a-a^2=0\\ \Leftrightarrow4a-1-a^2=0\\ \Leftrightarrow-a^2+4a-1=0\\ \Leftrightarrow a^2-4a+1=0\)

\(a=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4.1.1}}{2.1}\\ a=\dfrac{4\pm\sqrt{16-4}}{2}\\ a=\dfrac{4\pm\sqrt{12}}{2}\\ a=\dfrac{4\pm2\sqrt{3}}{2}\)

\(\left[{}\begin{matrix}a=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}\\a=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\end{matrix}\right.\)

Vậy......

Sr bạn cách làm mình hơi khó hiểu chút :v

a)ĐK: \(a\ge0\)

\(\Leftrightarrow\left(1+a\right)\left(1+\sqrt{6}\right)=\sqrt{6}\left(a+\sqrt{a}+1\right)\)

\(\Leftrightarrow1+\sqrt{6}+a+a\sqrt{6}=a\sqrt{6}+\sqrt{6a}+\sqrt{6}\)

\(\Leftrightarrow1+a=\sqrt{6a}\)

\(\Leftrightarrow a^2+2a+1-6a=0\)

\(\Leftrightarrow a^2-4a+1=0\)

\(\Leftrightarrow\left(a-2\right)^2-3=0\)

\(\Leftrightarrow\left(a-2\right)^2=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-2=\sqrt{3}\\a-2=-\sqrt{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\sqrt{3}+2\\a=2-\sqrt{3}\end{matrix}\right.\)

b)

MK sẽ chứng minh tương đương:

\(\Leftrightarrow\dfrac{1+a}{1+\sqrt{a}+a}-\dfrac{2}{3}>0\)

\(\Leftrightarrow\dfrac{a-2\sqrt{a}+1}{1+\sqrt{a}+a}>0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-1\right)^2}{1+\sqrt{a}+a}>0\)

Ta có:

\(\left(\sqrt{a}-1\right)^2\ge0\left(1\right)\)

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

Mà \(\left(\sqrt{a}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Leftrightarrow1+\sqrt{a}+a>0\left(2\right)\)

Từ (1), (2)

=>\(\dfrac{\left(\sqrt{a}-1\right)^2}{1+\sqrt{a}+a}>0\)

=>\(\dfrac{1+a}{1+\sqrt{a}+a}>\dfrac{2}{3}\)

Sửa đề: \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(a,C=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\left(a>0;a\ne1;a\ne4\right)\\ C=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,C\ge\dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}\ge0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}\ge0\\ \Leftrightarrow\sqrt{a}-4\ge0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a\ge16\)

Lời giải:

ĐKXĐ: $x>0; x\neq 4$
\(A=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{\sqrt{x}+2}\)

\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)

Hiển nhiên $B>0$

Với $x>0; x\neq 4\Rightarrow 3(\sqrt{x}+2)\geq 6$

$\Rightarrow B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}<3$

Vậy $0< B< 3$. $B$ nguyên $\Leftrightarrow B\in\left\{1;2\right\}$

$\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}\in\left\{1;2\right\}$

$\Leftrightarrow x\in\left\{\frac{64}{9}; \frac{1}{9}\right\}$ (tm)

a: ĐKXĐ: b>=0; b<>1

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

\(=\dfrac{1}{1-b}+\dfrac{b^2+1}{b^2-1}\)

\(=\dfrac{-b-1+b^2+1}{b^2-1}=\dfrac{b\left(b-1\right)}{\left(b-1\right)\left(b+1\right)}=\dfrac{b}{b+1}\)

b: B>1/3

=>B-1/3>0

=>b/b+1-1/3>0

=>(3b-b-1)/(3b+3)>0

=>2b-1>0

=>b>1/2

Có bài ngược của bài này, bạn đăng và đã có lời giải thì chỉ cần đảo lại đáp án là được.

\(E=\sqrt{x}+\dfrac{4}{\sqrt{x}}-2=\dfrac{4\sqrt{x}}{9}+\dfrac{4}{\sqrt{x}}+\dfrac{5}{9}.\sqrt{x}-2\)

\(E\ge2\sqrt{\dfrac{16\sqrt{x}}{9\sqrt{x}}}+\dfrac{5}{9}.\sqrt{9}-2=\dfrac{7}{3}\)

\(E_{min}=\dfrac{7}{3}\) khi \(x=9\)

\(F=3\sqrt{x}+\dfrac{1}{\sqrt{x}}+1=2\sqrt{x}+\dfrac{1}{\sqrt{x}}+\sqrt{x}+1\)

\(F\ge2\sqrt{\dfrac{2\sqrt{x}}{\sqrt{x}}}+1.\sqrt{\dfrac{1}{2}}+1=\dfrac{2+5\sqrt{2}}{2}\)

\(F_{min}=\dfrac{2+5\sqrt{2}}{2}\) khi \(x=\dfrac{1}{2}\)

\(a,P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\left(x\ge0;x\ne9\right)\\ P=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+8}{\sqrt{x}+1}\)

\(b,x=14-6\sqrt{5}=\left(3-\sqrt{5}\right)^2\)

Thay vào P:

\(P=\dfrac{14-6\sqrt{5}+8}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}=\dfrac{\left(4+\sqrt{5}\right)\left(22-6\sqrt{5}\right)}{11}=\dfrac{55-2\sqrt{5}}{11}\)

a) \(P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\left(đk:x\ge0,x\ne9\right)\)

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

\(=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-3x+x\sqrt{x}+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+8}{\sqrt{x}+1}\)

b) \(P=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\dfrac{22-6\sqrt{5}}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\dfrac{22-6\sqrt{5}}{3-\sqrt{5}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}\)