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\(=\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\cdot\left(\dfrac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{-2}\right)-\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\cdot\dfrac{4\sqrt{3}}{-2}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{2}-1\right)\left(2-\sqrt{6}\right)}{-2}+\dfrac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-1\right)}{1}\cdot\dfrac{-4\sqrt{3}}{2}\)
\(=\dfrac{2\sqrt{3}-3\sqrt{2}+2\sqrt{2}-2\sqrt{3}-2+\sqrt{6}+4\sqrt{3}\left(2-\sqrt{2}-\sqrt{6}+\sqrt{3}\right)}{-2}\)
\(=\dfrac{\sqrt{2}-2+\sqrt{6}+8\sqrt{3}-4\sqrt{6}-12\sqrt{2}+12}{-2}\)
\(=-\dfrac{-11\sqrt{2}+8\sqrt{3}-3\sqrt{6}+10}{2}\)
2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
\(=3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)
\(=\left(\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\right)\left(4-2\sqrt{3}\right)\)
\(=\left(\sqrt{6}+\sqrt{5}\right)\left(4-2\sqrt{3}\right)\)
\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)
Bài 2:
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(P=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2.\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(P=\left[\dfrac{\left(a-1\right)^2}{4a}\right].\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\sqrt{a}-1}{a-1}\right)\)
\(P=\dfrac{\left(a-1\right)^2}{4a}.\dfrac{2\sqrt{a}.\left(-2\right)}{a-1}\)
\(P=\dfrac{\left(a-1\right)^2\left(-4\sqrt{a}\right)}{4a.\left(a-1\right)}\)
\(P=\dfrac{\left(a-1\right).\left(-\sqrt{a}\right)}{a}=\dfrac{-a\sqrt{a}+\sqrt{a}}{a}\)
Bài 1:
\(A=\dfrac{2}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}\)\(A=\dfrac{2\sqrt{2}}{2}-\dfrac{1\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2-1}\)
\(A=\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{1}+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(A=\sqrt{2}-\sqrt{3}-\sqrt{2}+\sqrt{3}+1\)
\(A=1\)
a) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}\)
\(=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}\)
\(=1\)
b)
\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2\sqrt{6}}{6}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}\)
\(=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{6+3\sqrt{3}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}}{6\sqrt{3}}}\)
\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{6+3\sqrt{3}}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\left(-\sqrt{3}+2+\sqrt{3}\right)}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{12+6\sqrt{3}}}{-2\sqrt{3}}\)
\(=\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}}{-2\sqrt{3}}\)
\(=\dfrac{3+\sqrt{3}}{-2\sqrt{3}}\)
\(=-\dfrac{\left(3+\sqrt{3}\right)\sqrt{3}}{6}\)
\(=-\dfrac{3\sqrt{3}+3}{6}\)
\(=-\dfrac{3\left(\sqrt{3}+3\right)}{6}\)
\(=-\dfrac{\sqrt{3}+1}{2}\)
a: \(=\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
d: \(=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)
a: Ta có: \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)
\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
=-5+2
=-3
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
Bài 1:
ĐKXĐ: \(\dfrac{5}{x^2+6}>=0\)
=>\(x^2+6>0\)
mà \(x^2+6>=6>0\forall x\)
nên \(x\in R\)
Bài 2:
a: Sửa đề: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\cdot\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(=\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)
\(=2\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-3\sqrt{2}-1\)
b: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{6}\)
\(=\dfrac{\sqrt{12}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)
c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
d: \(\sqrt[3]{54}=\sqrt[3]{27\cdot2}=3\sqrt[3]{2}\)
e: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=\sqrt[3]{-27}=-3\)
f: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)
\(=\sqrt{2}+1-\sqrt{2}+1=2\)