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\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
\(
1)Q = \left( {\dfrac{1}{{y - \sqrt y }} + \dfrac{1}{{\sqrt y - 1}}} \right):\left( {\dfrac{{\sqrt y + 1}}{{y - 2\sqrt y + 1}}} \right)\\
Q = \left( {\dfrac{1}{{\sqrt y \left( {\sqrt y - 1} \right)}} + \dfrac{1}{{\sqrt y - 1}}} \right).\dfrac{{y - 2\sqrt y + 1}}{{\sqrt y + 1}}\\
Q = \dfrac{{1 + \sqrt y }}{{\sqrt y \left( {\sqrt y - 1} \right)}}.\dfrac{{{{\left( {\sqrt y - 1} \right)}^2}}}{{\sqrt y + 1}}\\
Q = \dfrac{{\sqrt y - 1}}{{\sqrt y }}
\)
b) Thay \(y=3-2\sqrt{2}\) vào biểu thức ta được:
\(\dfrac{{\sqrt {3 - 2\sqrt 2 } - 1}}{{\sqrt {3 - 2\sqrt 2 } }} = \dfrac{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} - 1}}{{\sqrt {{{\left( {1 - \sqrt 2 } \right)}^2}} }} = \dfrac{{ \sqrt 2 - 1-1}}{{\sqrt 2 -1}} \\= \dfrac{{\sqrt 2-2 }}{{ \sqrt 2 -1}} = \dfrac{{(\sqrt 2 -2)\left( { \sqrt 2+1 } \right)}}{{\left( { \sqrt 2-1 } \right)\left( {\sqrt 2+1 } \right)}} = - \sqrt 2 \)
\(2)B = \dfrac{{\sqrt y - 1}}{{{y^2} - y}}:\left( {\dfrac{1}{{\sqrt y }} - \dfrac{1}{{\sqrt y + 1}}} \right)\\ B = \dfrac{{\sqrt y - 1}}{{y\left( {y - 1} \right)}}:\dfrac{{\sqrt y + 1 - \sqrt y }}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ B = \dfrac{{\sqrt y - 1}}{{y\left( {\sqrt y - 1} \right)\left( {\sqrt y + 1} \right)}}:\dfrac{1}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ B = \dfrac{1}{{y\left( {\sqrt y + 1} \right)}}.\sqrt y \left( {\sqrt y + 1} \right)\\ B = \dfrac{{\sqrt y }}{y} \)
b) Thay \(y=3+2\sqrt{2}\) vào biểu thức ta được:
\(B = \dfrac{{\sqrt {3 + 2\sqrt 2 } }}{{3 + 2\sqrt 2 }} = \dfrac{{\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} }}{{3 + 2\sqrt 2 }} = \dfrac{{\left( {1 + \sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}} = 3 - 2\sqrt 2 + 3\sqrt 2 - 4 = - 1 + \sqrt 2 \)
Nhiều quá @@
\(a,\frac{\sqrt{108x^3}}{\sqrt{12x}}=\frac{\sqrt{36.3.x^3}}{\sqrt{3.4.x}}=\frac{6\sqrt{3}.\sqrt{x}^3}{2\sqrt{3}.\sqrt{x}}=3\sqrt{x}^2=3x\)
\(b,\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\frac{\sqrt{13}.\sqrt{x^4}.\sqrt{y^6}}{\sqrt{16.13}.\sqrt{x^6}.\sqrt{y^6}}=\frac{\sqrt{13}.x^2y^3}{4\sqrt{13}x^3y^3}=\frac{1}{4x}\)
\(c,\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}+\sqrt{y}\right)^2\)
\(=\frac{\sqrt{x}^3+\sqrt{y}^3}{\sqrt{x}+\sqrt{y}}-\left(x+2\sqrt{xy}+y\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x-2\sqrt{xy}-y\)
\(=x-\sqrt{xy}+y-x-2\sqrt{xy}-y=-3\sqrt{xy}\)
\(d,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đk chỗ này là \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge\sqrt{1}\Rightarrow x\ge1\)nhé
\(e,\frac{x-1}{\sqrt{y}-1}.\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}.\frac{y-2\sqrt{y}+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
A.2
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