2:

a: Sửa đề: \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

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

=>\(A>=2\cdot\sqrt{\sqrt{a^2+2}\cdot\dfrac{1}{\sqrt{a^2+2}}}=2\)

A=2 thì a^2+2=1

=>a^2=-1(loại)

=>A>2 với mọi a

b: \(\Leftrightarrow\sqrt{a}+\sqrt{b}< =\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\)

=>\(a\sqrt{a}+b\sqrt{b}>=a\sqrt{b}+b\sqrt{a}\)

=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)>=0\)

=>(căn a+căn b)(a-2*căn ab+b)>=0

=>(căn a+căn b)(căn a-căn b)^2>=0(luôn đúng)

1

ĐK: `x>1`

PT trở thành:

\(\sqrt{\dfrac{2x-3}{x-1}}=2\\ \Leftrightarrow\dfrac{2x-3}{x-1}=2^2=4\\ \Leftrightarrow4x-4-2x+3=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\left(KTM\right)\)

Vậy PT vô nghiệm.

b

ĐK: \(x\ge2\)

Đặt \(t=\sqrt{x-2}\) (\(t\ge0\))

=> \(x=t^2+2\)

PT trở thành: \(t^2+2-5t+2=0\)

\(\Leftrightarrow t^2-5t+4=0\)

nhẩm nghiệm: `a+b+c=0` (`1+(-5)+4=0`)

\(\Rightarrow\left\{{}\begin{matrix}t=1\left(nhận\right)\\t=4\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\x=18\left(TM\right)\end{matrix}\right.\)

\(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{5\sqrt{x}}{\sqrt{x}+3}=\frac{22}{x-9}\left(ĐK:x\ge0;x\ne9\right)\)

\(\Leftrightarrow\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}=\frac{22}{x-9}\)

\(\Rightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)=22\)

\(\Leftrightarrow x+5\sqrt{x}+6-5x+15\sqrt{x}=22\)

\(\Leftrightarrow-4x+20\sqrt{x}-16=0\)

\(\Leftrightarrow x-5\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}-4=0\\\sqrt{x}-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=16\left(tm\right)\\x=1\left(tm\right)\end{array}\right.\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{1;16\right\}\)

a, bình phương rồi phân tích là ra

b, nhân chéo rồi phá ngoặc

\(\sqrt{x^2-9}-5\sqrt{x+3}=0\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-5\sqrt{x+3}=0\)

ĐK: \(x+3\ge0\Leftrightarrow x\ge-3\) và  \(x-3\ge0\Leftrightarrow x\ge3\) suy ra điều kiện là X >=3

PT \(\Leftrightarrow\sqrt{\left(x+3\right)}\left(\sqrt{x+3}-5\right)=0\Leftrightarrow\sqrt{x+3}=0hoặc\left(\sqrt{x+3}-5\right)=0\)

+) \(\sqrt{x+3}=0\Leftrightarrow x=-3loai\)

+) \(\sqrt{x-3}-5=0\Leftrightarrow\sqrt{x-3}=5\Leftrightarrow x-3=25\Leftrightarrow x=28\)

Vậy x = 28

\(\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}+3}\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)Điều kiện x>=0

\(\Leftrightarrow x+\sqrt{x}-6=x-1\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\)

Vậy x = 25

1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x+2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy x=2 hoặc x=-1