Trục căn
\(\frac{1}{\sqrt[3]{3}+\sqrt[3]{2}}\)
Những câu hỏi liên quan
Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
trục căn thức ở mẫu :
\(\frac{1}{\sqrt{2\sqrt{3}-\sqrt{2}}.\sqrt{\sqrt{2}}+\sqrt{3}}\)
Trục căn thức ở mẫu
a)\(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
b)\(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
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b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
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trục căn thức ở mẫu:
\(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)
Ta có : \(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{12}+\sqrt{18}-\sqrt{24}-\sqrt{36}}{-6}\)\(=\frac{-\sqrt{12}-\sqrt{18}+\sqrt{24}+\sqrt{36}}{6}\)
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\(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)
trục căn thức ở mẫu
\(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)
\(=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}-2-\sqrt{6}\right)}{-6}\)
\(=2+\sqrt{6}-\sqrt{3}-\sqrt{2}\)
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\(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}\)
\(=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}-2-\sqrt{6}\right)}{-6}\)
\(=2+\sqrt{6}-\sqrt{3}-\sqrt{2}\)
Hông chắc !!!
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trục căn thức ở mẫu: \(\frac{1}{1+3\sqrt[3]{2}-2\sqrt[3]{4}}\)
trục căn ở mẫu \(\frac{1}{\sqrt[3]{2}+\sqrt{ }3}\)
\(\frac{1}{\sqrt[3]{2}+\sqrt{3}}=\frac{\left(\sqrt[3]{2}\right)^2-\sqrt[3]{2}\cdot\sqrt{3}+3}{2+3\sqrt{3}}=\frac{\left(\left(\sqrt[3]{2}\right)^2-\sqrt[3]{2}\cdot\sqrt{3}+3\right)\left(2-3\sqrt{3}\right)}{4-27}\)
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trục căn thức ở mẫu :
a,frac{3}{sqrt{5}};frac{2sqrt{3}}{sqrt{2}};frac{a}{sqrt{b}};frac{x+1}{sqrt{x^2-1}}
b,frac{1}{sqrt{3}+sqrt{2}};frac{2}{2-sqrt{3}};frac{sqrt{2}+1}{sqrt{2}-1};frac{3sqrt{2}}{sqrt{3}+1}
c,frac{1}{1+sqrt{2}+sqrt{3}}
d,frac{1}{sqrt{2sqrt{3}-sqrt{2}}.sqrt{2}.sqrt{sqrt{2}+sqrt{3}}}
Đọc tiếp
trục căn thức ở mẫu :
a,\(\frac{3}{\sqrt{5}};\frac{2\sqrt{3}}{\sqrt{2}};\frac{a}{\sqrt{b}};\frac{x+1}{\sqrt{x^2-1}}\)
b,\(\frac{1}{\sqrt{3}+\sqrt{2}};\frac{2}{2-\sqrt{3}};\frac{\sqrt{2}+1}{\sqrt{2}-1};\frac{3\sqrt{2}}{\sqrt{3}+1}\)
c,\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
d,\(\frac{1}{\sqrt{2\sqrt{3}-\sqrt{2}}.\sqrt{2}.\sqrt{\sqrt{2}+\sqrt{3}}}\)
a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)
\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)
\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)
\(\frac{x+1}{\sqrt{x^2-1}}=\frac{\left(x+1\right)\left(\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-1}\right)\left(\sqrt{x^2-1}\right)}\) = \(\frac{\left(\sqrt{x^2-1}\right)\left(x+1\right)}{x^2-1}\)
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câu c chắc là như này
\(\frac{1}{1+\sqrt{2}+\sqrt{3}}=1+\frac{1}{\sqrt{2}+\sqrt{3}}\) = \(1+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(1+\frac{\sqrt{2}-\sqrt{3}}{2-3}=1+\frac{\sqrt{2}-\sqrt{3}}{-1}\) = \(1-\sqrt{2}+\sqrt{3}\)
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trục căn thức ở mẫu:
a) \(\frac{\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}\)
b) \(\frac{\sqrt{a+3}-\sqrt{a-3}}{\sqrt{a+3}+\sqrt{a-3}}\left(a\ge3\right)\)