a,Với \(a>0;a\ne1\)

 \(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

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

b, Ta có : \(1=\frac{a+\sqrt{a}}{a+\sqrt{a}}\)mà \(a-1=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\)

\(a+\sqrt{a}=\sqrt{a}\left(\sqrt{a}+1\right)\)vì \(\sqrt{a}-1< \sqrt{a}\)

Vậy \(\frac{a-1}{a+\sqrt{a}}< 1\)hay \(M< 1\)

a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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

Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)

\(\Leftrightarrow\sqrt{x}+1\ge1\)

\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)

\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)

\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)

Dấu "=" xảy ra khi x=0

Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0

Mk lấy KQ rút gọn của 💋Amanda💋 nha!

d/ \(M=\frac{\sqrt{a}-2}{\sqrt{a}+1}=\frac{\sqrt{a}+1-3}{\sqrt{a}+1}=1-\frac{3}{\sqrt{a}+1}\)

Có \(\sqrt{a}+1\ge1\Rightarrow\frac{3}{\sqrt{a}+1}\le3\)

\(\Rightarrow1-\frac{3}{\sqrt{a}+1}\ge1-3=-2\)

"="\(\Leftrightarrow a=0\)

a/ Điều kiện \(\hept{\begin{cases}a\ge0\\a\ne\frac{1}{9}\end{cases}}\) \(\Rightarrow0\le a\ne\frac{1}{9}\)

b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(a-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)

\(=\frac{2\sqrt{a}\left(1-3\sqrt{a}\right)+\left(\sqrt{a}-2\right)\left(1+3\sqrt{a}\right)+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}:\left(\frac{3a\sqrt{a}-2\sqrt{a}+6-a}{3\sqrt{a}-1}\right)\)

\(=\frac{2\sqrt{a}-6a+\sqrt{a}+3a-2-6\sqrt{a}+5\sqrt{a}+3}{\left(1-3\sqrt{a}\right)\left(1+3\sqrt{a}\right)}.\left(\frac{3\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\right)\)

\(=\frac{3a-2\sqrt{a}-1}{1+3\sqrt{a}}.\frac{1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)

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

\(=\frac{\sqrt{a}-1}{3a\sqrt{a}-2\sqrt{a}+6-a}\)

Hình như đề sai rồi bạn :(

a/ Điều kiện xác định : \(\hept{\begin{cases}a\ge0\\a\ne9\end{cases}\Leftrightarrow}0\le a\ne9\)

b/ \(M=\left(\frac{2\sqrt{a}}{3\sqrt{a}+1}+\frac{\sqrt{a}-2}{1-3\sqrt{a}}-\frac{5\sqrt{a}+3}{9a-1}\right):\left(1-\frac{2\sqrt{a}-6}{3\sqrt{a}-1}\right)\)

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

\(=\frac{6a-2\sqrt{a}+6\sqrt{a}+2-3a-\sqrt{a}-5\sqrt{a}-3}{\left(3\sqrt{a}+1\right)\left(3\sqrt{a}-1\right)}.\frac{3\sqrt{a}-1}{\sqrt{a}+5}\)

\(=\frac{3a-2\sqrt{a}-1}{3\sqrt{a}+1}.\frac{1}{\sqrt{a}+5}\)

\(=\frac{\left(3\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(3\sqrt{a}+1\right)\left(\sqrt{a}+5\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+5}\)

c/ \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\) thay vào M được

\(\frac{\sqrt{5}-2-1}{\sqrt{5}-2+5}=\frac{\sqrt{5}-3}{\sqrt{5}+3}=\frac{-7+3\sqrt{5}}{2}\)

d/ \(M=\frac{\sqrt{a}-1}{\sqrt{a}+5}=\frac{\sqrt{a}+5-6}{\sqrt{a}+5}=1-\frac{6}{\sqrt{a}+5}\)

Với mọi \(0\le a\ne9\) thì ta luôn có \(\sqrt{a}+5\ge5\Leftrightarrow\frac{6}{\sqrt{a}+5}\le\frac{6}{5}\Leftrightarrow-\frac{6}{\sqrt{a}+5}\ge-\frac{6}{5}\Leftrightarrow1-\frac{6}{\sqrt{a}+5}\ge1-\frac{6}{5}\)

\(\Rightarrow M\ge-\frac{1}{5}\)

Đẳng thức xảy ra khi a = 0

Vậy giá trị nhỏ nhất của M bằng \(-\frac{1}{5}\) khi a = 0

Điều kiện \(a\ne\frac{1}{9}\) nhé!

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

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

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

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

b) PT có nghiệm <=> x>0

<=>\(\sqrt{x}>0\)

<=> \(\sqrt{x}-1>-1\)

<=> x>-1

Đậu mé.

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