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\(Q=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)
DK: \(a,b\ge0\)do \(Q=2\sqrt{b}\)nên Q ko phụ thuộc vào giá trị của a
a,ĐK:\(a>0;b>0;a\ne b\)
b,\(A=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\\ A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\\ A=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=0\)
Vậy khi A có nghĩa thì A không phụ thuộc vào a
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
\(A=\frac{a+b+2\sqrt{ab}-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(A=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}+\sqrt{b}\right)\)
\(A=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
\( Q = \dfrac{{{{\left( {\dfrac{{a - b}}{{\sqrt a + \sqrt b }}} \right)}^3} + 2a\sqrt a + b\sqrt b }}{{3{a^2} + 3b\sqrt {ab} }} + \dfrac{{\sqrt {ab} - a}}{{a\sqrt a - b\sqrt a }}\\ Q = \dfrac{{{{\left[ {\dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a + \sqrt b }}} \right]}^3} + 2a\sqrt a + b\sqrt b }}{{3\left( {{a^2} + b\sqrt {ab} } \right)}} + \dfrac{{\sqrt a \left( {\sqrt b - \sqrt a } \right)}}{{\sqrt a \left( {a - b} \right)}}\\ Q = \dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^3} + 2a\sqrt a + b\sqrt b }}{{3\sqrt a \left( {a\sqrt a + b\sqrt b } \right)}} + \dfrac{{ - \left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}\\ Q = \dfrac{1}{{\sqrt a + \sqrt b }} + \dfrac{{ - 1}}{{\sqrt a + \sqrt b }} = 0 \)
Vậy Q không phụ thuộc vào a,b
Bài làm:
a) đkxđ: \(a\ne1;a>0\)
b) Ta có:
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right)\div\frac{\sqrt{a}+1}{a}\)
\(A=\left[\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right].\frac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}\)
\(A=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(A=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)
\(A=\left(\sqrt{a}-1\right)\sqrt{a}\)
\(A=a-\sqrt{a}\)
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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
\(A=\left(\frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3}\right)\left(1-\frac{3}{\sqrt{a}}\right)\) \(đk:a>0;a\ne9\)
\(=\frac{\sqrt{a}+3+\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}.\frac{\sqrt{a}-3}{\sqrt{a}}\)
\(=\frac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
\(=\frac{2}{\sqrt{a}+3}\)
\(đk:a>0;a\ne9\)
\(A>\frac{1}{2}=>\frac{2}{\sqrt{a}+3}>\frac{1}{2}\)
\(=>4>\sqrt{a}+3\)
\(< =>\sqrt{a}>1\)
\(< =>a=1\)