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Lời giải:
Yêu cầu 1:
\(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-(\sqrt{5}+3)=\frac{\sqrt{5}(\sqrt{5}+3)}{\sqrt{5}}+\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}-(\sqrt{5}+3)\)
\(=\sqrt{5}+3+\sqrt{3}-(\sqrt{5}+3)=\sqrt{3}\) (đpcm)
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Yêu cầu 2:
\(P=a-\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{(\sqrt{a}-\sqrt{a-1})(\sqrt{a}+\sqrt{a-1})}=a-\frac{2\sqrt{a-1}}{a-(a-1)}=a-2\sqrt{a-1}\)
\(=(a-1)-2\sqrt{a-1}+1=(\sqrt{a-1}-1)^2\geq 0\) với mọi $a\geq 1$
Ta có đpcm.
Câu 2:
\( P = \dfrac{{a - b}}{{\sqrt a + \sqrt b }} + \dfrac{{a\sqrt a - b\sqrt b }}{{a + b + \sqrt {ab} }}\\ P = \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\left( {\sqrt a + \sqrt b } \right)}} + \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}{{a + b + \sqrt {ab} }}\\ P= \sqrt a - \sqrt b + \sqrt a - \sqrt b \\ P = 2\sqrt a - 2\sqrt b \)
Câu 1:
\(a)\left( {3\sqrt {\dfrac{3}{5}} - \sqrt {\dfrac{5}{3}} + \sqrt 5 } \right)2\sqrt 5 + \dfrac{2}{3}\sqrt {75} \\
= 6\sqrt {\dfrac{{15}}{5}} - 2\sqrt {\dfrac{{25}}{3}} + 10 + \dfrac{{10\sqrt 3 }}{3}\\
= 6\sqrt 3 - \dfrac{{10}}{{\sqrt 3 }} + 10 + \dfrac{{10\sqrt 3 }}{3}\\
= 6\sqrt 3 - \dfrac{{10\sqrt 3 }}{3} + 10 + \dfrac{{10\sqrt 3 }}{3}\\
= 6\sqrt 3 + 10\\
b){\left( {\sqrt 3 - 1} \right)^2} - \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} + \sqrt {{{\left( { - 3} \right)}^2}.3} \\
= 3 - 2\sqrt 3 + 1 - \sqrt 3 + 1 + \sqrt {{3^3}} \\
= 5 - 3\sqrt 3 + 3\sqrt 3 \\
= 5\)
\(a,\left(3\sqrt{\frac{3}{5}}-\sqrt{\frac{5}{3}}+\sqrt{5}\right)2\sqrt{5}+\frac{2}{3}\sqrt{75}\)
\(=6\sqrt{3}-\frac{10\sqrt{3}}{3}+10+\frac{10\sqrt{3}}{3}\)
\(=6\sqrt{3}+10\)
\(b,\left(\sqrt{3}-1\right)^2-\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{\left(-3\right)^2.3}\)
\(=\left(\sqrt{3}^2-2.\sqrt{3}.1+1^2\right)-|1-\sqrt{3}|+\sqrt{27}\)
\(=4-2\sqrt{3}-\sqrt{3}+1+3\sqrt{3}\)
\(=5\)
\(P=\frac{a-b}{\sqrt{a}+\sqrt{b}}+\frac{a\sqrt{a}-b\sqrt{b}}{a+b+\sqrt{ab}}\left(a\ge0;b\ge0;a\ne b\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{a+b+\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}+\sqrt{a}-\sqrt{b}\)
\(=2\sqrt{a}-2\sqrt{b}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)
\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
\(c,\left(3+\frac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3-\frac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)=\left(3+\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\right)\left(3-\frac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)
\(=\left(3+\sqrt{a}\right)\left(3-\sqrt{a}\right)=3^2-\sqrt{a}^2=3-a\)
\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)
