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ĐKXĐ:\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a>0\\b>0\end{matrix}\right.\\\left\{{}\begin{matrix}a< 0\\b< 0\end{matrix}\right.\end{matrix}\right.\) và \(a\ne b\)
a)\(N=\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\)
\(=\frac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\frac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\frac{a+b}{\sqrt{ab}}\)
\(=\frac{a\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)+b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(b-a\right)}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}-\sqrt{a}\right)}\)
\(=\frac{a\sqrt{ab}-a^2+b^2+b\sqrt{ab}+b^2-a^2}{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}-\sqrt{a}\right)}\)
\(=\frac{\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}\)
\(=\frac{a+b}{b-a}\)
b)\(a=\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(b=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Khi đó, \(N=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}-1-\sqrt{3}-1}=\frac{2\sqrt{3}}{-2}=-\sqrt{3}\)
c)\(\frac{a}{b}=\frac{a+1}{b+5}\Leftrightarrow ab+5a=ab+b\Leftrightarrow5a=b\)
Thay vào N ta được:
\(N=\frac{a+5a}{5a-a}=\frac{6a}{4a}=\frac{3}{2}\)(luôn không đổi)
Cho \(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}\)
a, Rút gọn Q
B, Chứng minh Q=\(\frac{b+81}{b-81}\)thì \(\frac{b}{a}\)là một số nguyên chia hết cho 3
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{a}\left(\sqrt{b}+2\right)-3\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2\sqrt{a}+3\sqrt{b}}{\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}-\frac{6-\sqrt{ab}}{\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}-\frac{\left(\sqrt{a}-3\right)\left(6-\sqrt{ab}\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(2\sqrt{a}+3\sqrt{b}\right)\left(\sqrt{a}+3\right)-\left(\sqrt{a}-3\right)\left(6-\sqrt{ab}\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2a+6\sqrt{a}+3\sqrt{ab}+9\sqrt{b}-6\sqrt{a}+a\sqrt{b}+18-3\sqrt{ab}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{2a+9\sqrt{b}+a\sqrt{b}+18}{\left(a-9\right)\left(\sqrt{b}+2\right)}\)
\(Q=\frac{\left(a+9\right)\left(\sqrt{b}+2\right)}{\left(a-9\right)\left(\sqrt{b}+2\right)}=\frac{a+9}{a-9}\)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)