Trục \(\dfrac{1}{1+\sqrt{2}+\sqrt{3}}\)
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24 tháng 10 2021
\(\dfrac{-2}{\sqrt{3}-1}=\dfrac{-2\left(\sqrt{3}+1\right)}{2}=-\sqrt{3}-1\\ \dfrac{\sqrt{5}}{\sqrt{7}-3}=\dfrac{-\sqrt{5}\left(\sqrt{7}+3\right)}{2}\\ \dfrac{3\sqrt{3}-2}{1-2\sqrt{3}}=\dfrac{\left(3\sqrt{3}-2\right)\left(1+2\sqrt{3}\right)}{-11}=\dfrac{\sqrt{3}-16}{11}\\ \dfrac{14}{\sqrt{10}+\sqrt{3}}=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{7}=2\sqrt{10}-2\sqrt{3}\)
6 tháng 7 2021
a) Ta có: \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-5-2\sqrt{6}}\)
\(=\dfrac{-\sqrt{5}-\sqrt{3}-\sqrt{2}}{2\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)}{12}\)
b) Ta có: \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)
\(=\dfrac{2\left(-1-\sqrt{2}-\sqrt{3}\right)}{\left(-1-\sqrt{2}\right)^2-3}\)
\(=\dfrac{\left(-1-\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{-\sqrt{2}-2-\sqrt{6}}{2}\)
AH
Akai Haruma
Giáo viên
4 tháng 7 2021
Lời giải:
a.
\(\frac{1}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\frac{\sqrt{5}+\sqrt{3}}{5-3}=\frac{\sqrt{5}+\sqrt{3}}{2}\)
b.
\(=\frac{2[(\sqrt{3}-(\sqrt{2}-1)]}{[(\sqrt{3}+(\sqrt{2}-1)][\sqrt{3}-(\sqrt{2}-1)]}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{3-(\sqrt{2}-1)^2}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{2\sqrt{2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+1}{\sqrt{2}}=\frac{\sqrt{6}-2+\sqrt{2}}{2}\)
c.
\(=\frac{5(\sqrt[3]{2^2}-3\sqrt[3]{2}+3^2)}{(\sqrt[3]{2})^3+3^3}=\frac{5(\sqrt[3]{4}+3\sqrt[3]{2}+9)}{29}\)
16 tháng 5 2023
\(B=\dfrac{\left(1+\sqrt{5}\right)\left(2+\sqrt{5}\right)}{-1}=-2-3\sqrt{5}-5=-7-3\sqrt{5}\)
\(C=\dfrac{5\sqrt{x}-x}{2x}\)
\(D=\dfrac{\left(\sqrt{a}+1\right)\left(2\sqrt{a}+1\right)}{4a-1}\)
\(E=\dfrac{15}{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{\sqrt{15}}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{75}+\sqrt{45}}{2}\)
AT
16 tháng 7 2021
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}-\sqrt{2}-1\right)}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}-1}{3-\left(\sqrt{2}+1\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}-1}{-2\sqrt{2}}=\dfrac{\left(\sqrt{3}-\sqrt{2}-1\right)\sqrt{2}}{-2\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}-2-\sqrt{2}}{-4}\)
\(=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)
17 tháng 7 2021
\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)
\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)
6 tháng 7 2021
a) Ta có: \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{7}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{5}-\sqrt{3}\right)^2-7}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)}{1-2\sqrt{15}}\)
\(=\dfrac{7\left(\sqrt{5}-\sqrt{3}-\sqrt{7}\right)\left(1+2\sqrt{15}\right)}{1-60}\)
\(=\dfrac{-7\left(\sqrt{5}+10\sqrt{3}-\sqrt{3}-6\sqrt{5}-\sqrt{7}-2\sqrt{105}\right)}{59}\)
\(=\dfrac{-7\left(-5\sqrt{5}+9\sqrt{3}-\sqrt{7}-2\sqrt{105}\right)}{59}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Bài 1:
a.
\(\frac{1}{2\sqrt{2}-3\sqrt{3}}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2}-3\sqrt{3})(2\sqrt{2}+3\sqrt{3})}=\frac{2\sqrt{2}+3\sqrt{3}}{(2\sqrt{2})^2-(3\sqrt{3})^2}=\frac{2\sqrt{2}+3\sqrt{3}}{-19}\)
b.
\(=\sqrt{\frac{(3-\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5})^2}{3^2-5}}=\sqrt{\frac{(3-\sqrt{5})^2}{4}}=\sqrt{(\frac{3-\sqrt{5}}{2})^2}=|\frac{3-\sqrt{5}}{2}|=\frac{3-\sqrt{5}}{2}\)
AH
Akai Haruma
Giáo viên
19 tháng 7 2021
Bài 2.
a.
\(=\frac{\sqrt{8}(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{2\sqrt{2}(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{2}(\sqrt{5}+\sqrt{3})=\sqrt{10}+\sqrt{6}\)
b.
\(=\sqrt{\frac{(2-\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}=\sqrt{\frac{(2-\sqrt{3})^2}{2^2-3}}=\sqrt{(2-\sqrt{3})^2}=|2-\sqrt{3}|=2-\sqrt{3}\)
AT
4 tháng 7 2021
\(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1+1}}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}+1\right|\)
\(=\sqrt{x-1}-1+\sqrt{x-1}+1\left(x\ge2\right)=2\sqrt{x-1}\)
a) \(\dfrac{1}{\sqrt{5}+\sqrt{7}}=\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\dfrac{\sqrt{7}-\sqrt{5}}{2}\)
c) \(\dfrac{7}{\sqrt{5}-\sqrt{3}+\sqrt{5}}=\dfrac{7}{2\sqrt{5}-\sqrt{3}}=\dfrac{7\left(2\sqrt{5}+\sqrt{3}\right)}{\left(2\sqrt{5}+\sqrt{3}\right)\left(2\sqrt{5}-\sqrt{3}\right)}\)
\(=\dfrac{14\sqrt{5}+7\sqrt{3}}{17}\)
\(\dfrac{1}{1+\sqrt{2}+\sqrt{3}}=\dfrac{1+\sqrt{2}-\sqrt{3}}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)}\)
\(=\dfrac{1+\sqrt{2}-\sqrt{3}}{\left(1+\sqrt{2}\right)^2-3}=\dfrac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\left(1+\sqrt{2}-\sqrt{3}\right)}{4}=\dfrac{\sqrt{2}+2-\sqrt{6}}{4}\)