Tính:
a) \(\sqrt[3]{{ - 125}}\);
b) \(\sqrt[4]{{\frac{1}{{81}}}}.\)
K
Khách
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Những câu hỏi liên quan
12 tháng 7 2021
Bài 1:
a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)
b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)
Bài 2:
a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)
b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)
c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)
\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)
\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)
\(=\dfrac{6}{35}\)
24 tháng 11 2021
\(a,=6\sqrt{2}-3-6\sqrt{2}=-3\\ b,=12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}=6\sqrt{3}+3\sqrt{5}\\ c,=\sqrt{3}-1-\sqrt{3}=-1\\ d,=\sqrt{6}-\dfrac{5\left(\sqrt{6}+1\right)}{5}=\sqrt{6}-\sqrt{6}-1=-1\)
RK
23 tháng 6 2021
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
RK
23 tháng 6 2021
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
25 tháng 6 2021
`A=sqrt{3-2sqrt2}-sqrt{3+2sqrt2}`
`=sqrt{2-2sqrt2+1}-sqrt{2+2sqrt2+1}`
`=sqrt{(sqrt2-1)^2}-sqrt{(sqrt2+1)^2}`
`=|sqrt2-1|-|\sqrt2+1|`
`=sqrt2-1-sqrt2-1=-2`
25 tháng 6 2021
Ta có: \(A=\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2}-1-\sqrt{2}-1\)
=-2
B
1
15 tháng 8 2023
a: \(=\sqrt[3]{\dfrac{5}{625}}=\sqrt[3]{\dfrac{1}{125}}=\dfrac{1}{5}\)
b: \(=\sqrt[5]{\left(-\sqrt{5}\right)^5}=-\sqrt{5}\)
2 tháng 9 2023
\(A=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{99}+\sqrt{100}\)
=10-1
=9
30 tháng 6 2021
a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)
\(=-3\cdot4\cdot3\sqrt{10}\)
\(=-36\sqrt{10}\)
b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)
\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)
\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)
\(=15\sqrt{3}\)
c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)
\(=4\cdot3-4-2\cdot2\)
\(=12-4-4=4\)
26 tháng 9 2021
a)\(\sqrt{\dfrac{2}{2-\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)\(=\sqrt{2\left(2+\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\sqrt{2^2.6}+\dfrac{2\sqrt{24}}{8}+\dfrac{1}{6}\)
\(=\dfrac{\sqrt{6}}{3}-2\sqrt{6}+\dfrac{\sqrt{2^2.6}}{4}+\dfrac{1}{6}=\dfrac{-5\sqrt{6}}{3}+\dfrac{2\sqrt{6}}{4}+\dfrac{1}{6}\)
\(=\dfrac{-7\sqrt{6}}{6}+\dfrac{1}{6}\)
26 tháng 9 2021
Nhầm xíu
\(\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\left|\sqrt{3}-1\right|=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
a: \(\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)^3}=-5\)
b: \(\sqrt[4]{\dfrac{1}{81}}=\sqrt[4]{\left(\dfrac{1}{3}\right)^4}=\dfrac{1}{3}\)
a) \(\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)^3}=-5\)
b) \(\sqrt[4]{\dfrac{1}{81}}=\sqrt[4]{\left(\dfrac{1}{3}\right)^4}=\dfrac{1}{3}\)