a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)

\(M=\sqrt{3}-\sqrt{3}+1\)

\(M=1\)

b) Ta có:

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

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

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

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

\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Theo đề ta có: \(M=2N\)

Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)

\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)

\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)

\(\Leftrightarrow\sqrt{a}=2\)

\(\Leftrightarrow a=4\left(tm\right)\)

a ĐK \(a>0\)và \(a\ne1\)

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

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

b. Ta có \(M-1=\frac{\sqrt{a}-1}{\sqrt{a}}-1=\frac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=\frac{-1}{\sqrt{a}}< 0\)

Vậy \(M< 1\)

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 -.- 

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

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

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

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

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

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

1: Khi x=9 thì \(A=\dfrac{3+1}{3-1}=\dfrac{4}{2}=2\)

2:

a: \(P=\left(\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

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

b: \(2P=2\sqrt{x}+5\)

=>\(P=\sqrt{x}+\dfrac{5}{2}\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}=\dfrac{2\sqrt{x}+5}{2}\)

=>\(\sqrt{x}\left(2\sqrt{x}+5\right)=2\sqrt{x}+2\)

=>\(2x+3\sqrt{x}-2=0\)

=>\(\left(\sqrt{x}+2\right)\left(2\sqrt{x}-1\right)=0\)

=>\(2\sqrt{x}-1=0\)

=>x=1/4

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

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

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

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

b) PT có nghiệm <=> x>0

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

<=> \(\sqrt{x}-1>-1\)

<=> x>-1

Đậu mé.

Yessssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss

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

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

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

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

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

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

Tại \(x=4+2\sqrt{3}\): \(\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

\(A=\frac{\sqrt{3}}{4+2\sqrt{3}+\sqrt{3}+1}=\frac{\sqrt{3}}{5+3\sqrt{3}}\)

\(A-1=\frac{\sqrt{x}-1}{x+\sqrt{x}}-1=\frac{-1-x}{x+\sqrt{x}}< 0\)do \(x>0\).

Vậy \(A< 1\).