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) ĐKXĐ : \(a>0;a\ne1\)

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

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

\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)

\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)

b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)

\(\Rightarrow0< a< \frac{4}{25}\)

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

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

P = \(\frac{\left(a-1\right)^2}{4a}\cdot\frac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{a-1}\)

P = \(\frac{a-1}{4\sqrt{a}^2}\cdot\left(-4\sqrt{a}\right)\)

P = \(\frac{1-a}{\sqrt{a}}\)

b) với x > 0 và x khác 1

P < 0 => \(\frac{1-a}{\sqrt{a}}< 0\)

Do \(\sqrt{a}>0\) => 1 - a < 0 => a > 1

Vậy S = {a|a > 1}

Có 1 kiểu hơi khác Conan 1 tí -.-

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

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

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

Vậy \(P=\frac{1-a}{\sqrt{a}}\)với a > 0 và \(a\ne1\)

b) Do a > 0 và a khác 1 nên P < 0 khi và chỉ khi :

\(\frac{1-a}{\sqrt{a}}< 0\Leftrightarrow1-a< 0\Leftrightarrow a>1\)

ơ, t làm t vừa thắc mắc nhá. \(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)chỗ này lược đc hay mấy you làm kiểu j vậy -.- 

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

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

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

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

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

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

a) \(P=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

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

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

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

b)Ta có a>0 do đó: \(P=a-\sqrt{a}\ge0\)

Dấu "=" xảy ra khi a=1

c) Ta thấy \(P\ge0\)

=>P²\(\ge\)P

=>P\(\ge\)\(\sqrt{P}\)

ĐKXĐ: \(a>0\)

a/ \(P=\frac{\sqrt{a}\left(\sqrt{a}^3+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

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

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

         \(=a-\sqrt{a}\)

b/ Ta có: \(\hept{\begin{cases}a>0\\\sqrt{a}\ge0\end{cases}\Rightarrow a-\sqrt{a}\ge0}\) 

    MinP = 0 khi \(\sqrt{a}=0\Rightarrow a=0\)

c/ \(P\ge\sqrt{P}\)

a)

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

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

\(=\frac{4a\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}=\frac{2}{a-1}\)

b) \(\frac{2}{a-1}=a\Rightarrow a^2-a-2=0\)

Ta có: 1+1+(-2)=0, nên pt có 2 nghiệm a₁=-1<0 (không thỏa mãn đk)=> loại

a₂=2(thỏa mãn đk)=> chọn

Vậy a=2 thì P=a