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\(A=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}.\)
\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\frac{4\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)
\(B=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\)\(\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2.\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)
= \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{2}{a-1}\)
b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1
=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 }
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`)
\(\Rightarrow\left\{{}\begin{matrix}t=1\left(nhận\right)\\t=4\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{x-2}=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\x=18\left(TM\right)\end{matrix}\right.\)
Ta có : \(P=\frac{a^3-a-2b-\frac{b^2}{a}}{\left(\frac{1}{\sqrt{a}}-\sqrt{\frac{1}{a}+\frac{b}{a^2}}\right)\left(\sqrt{a}+\sqrt{a+b}\right)}:\left(\frac{a^3+a^2+ab+a^2b}{a^2-b^2}+\frac{b}{a-b}\right)\)
=> \(P=\frac{\frac{a^4}{a}-\frac{a^2}{a}-\frac{2ab}{a}-\frac{b^2}{a}}{\left(\frac{1}{\sqrt{a}}-\sqrt{\frac{1}{a}+\frac{b}{a^2}}\right)\left(\sqrt{a}+\sqrt{a+b}\right)}:\left(\frac{a^2\left(a+1\right)+ab\left(a+1\right)}{\left(a-b\right)\left(a+b\right)}+\frac{b}{a-b}\right)\)
=> \(P=\frac{\frac{a^4-a^2-2ab-b^2}{a}}{\frac{\sqrt{a}}{\sqrt{a}}-\sqrt{a\left(\frac{1}{a}+\frac{b}{a^2}\right)}+\sqrt{\frac{a+b}{a}}-\sqrt{\left(a+b\right)\left(\frac{1}{a}+\frac{b}{a^2}\right)}}:\left(\frac{a\left(a+b\right)\left(a+1\right)}{\left(a-b\right)\left(a+b\right)}+\frac{b}{a-b}\right)\)
=> \(P=\frac{\frac{a^4-\left(a^2+2ab+b^2\right)}{a}}{1-\sqrt{\frac{a}{a}+\frac{ab}{a^2}}+\sqrt{\frac{a+b}{a}}-\sqrt{\frac{a}{a}+\frac{b}{a}+\frac{ab}{a^2}+\frac{b^2}{a^2}}}:\left(\frac{a\left(a+1\right)+b}{a-b}\right)\)
=> \(P=\frac{\frac{a^4-\left(a^2+2ab+b^2\right)}{a}}{1-\sqrt{1+\frac{b}{a}}+\sqrt{\frac{a+b}{a}}-\sqrt{1+\frac{2b}{a}+\frac{b^2}{a^2}}}:\left(\frac{a\left(a+1\right)+b}{a-b}\right)\)
=> \(P=\frac{\frac{a^4-\left(a+b\right)^2}{a}\left(a-b\right)}{\left(1-\sqrt{1+\frac{b}{a}}+\sqrt{\frac{a+b}{a}}-\left(\frac{b}{a}+1\right)\right)\left(a\left(a+1\right)+b\right)}\)
=> \(P=\frac{\frac{\left(a^2-a-b\right)\left(a^2+a+b\right)\left(a-b\right)}{a}}{\left(1-\frac{b}{a}-1\right)\left(a\left(a+1\right)+b\right)}\)\(=\frac{\frac{\left(a^2-a-b\right)\left(a^2+a+b\right)\left(a-b\right)}{a}}{\frac{b\left(a^2+a+b\right)}{a}}\)\(=\frac{\left(a^2-a-b\right)\left(a^2+a+b\right)\left(a-b\right)}{b\left(a^2+a+b\right)}\)
=> \(P=\frac{\left(a^2-a-b\right)\left(a-b\right)}{b}\)
- Thay a = 23, b = 22 vào biểu thức trên ta được :
\(P=\frac{\left(23^2-23-22\right)\left(23-22\right)}{22}=22\)