TÍNH \(^{\sqrt[3]{162}-\sqrt[3]{48}-\sqrt[3]{6}}\)
Những câu hỏi liên quan
\(\sqrt[3]{162}-\sqrt[3]{48}-\sqrt[3]{6}-\sqrt[3]{-0,008}+\sqrt[3]{\dfrac{8}{125}}\)
\(=3\sqrt[3]{6}-2\sqrt[3]{6}-\sqrt[3]{6}-\left(-0,2\right)+\dfrac{2}{5}\\ =\dfrac{2}{5}+0,2=\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{3}{5}\)
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Tính:
\(\sqrt[3]{48}-\sqrt[3]{384}+\sqrt[3]{162}\)
thực hiện phép tính
a, \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
b, \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
c, \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
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\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).
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\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=\dfrac{\left(\sqrt{2-\sqrt{3}}\right)^2+\left(\sqrt{2+\sqrt{3}}\right)^2}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}=\dfrac{4}{1}=4\)
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Tính chính xác kết quả của phép tính sau:\(A=5\sqrt[3]{6\sqrt{32}}-3\sqrt[3]{9\sqrt{162}}-11\sqrt[6]{18}+2\sqrt[3]{75\sqrt{50}}-7\sqrt[3]{45\sqrt{27}}\)
Thực hiện phép tính:
a) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
b) \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
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rút gọn
d,\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\) e,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) f,\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
=căn 5-1/2*căn 5
=1/2*căn 5
e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)
f:=2+căn 3+căn 2-2-căn 3=căn 2
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Thực hện phép tínha, frac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}-frac{sqrt{5}+sqrt{27}}{sqrt{30}+sqrt{162}}b,frac{sqrt{3-sqrt{5}}.left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}}c, sqrt{4+sqrt{10+2sqrt{5}}}+sqrt{4-sqrt{10+2sqrt{5}}}d, left(frac{sqrt{6}-sqrt{2}}{1-sqrt{3}}-frac{5}{sqrt{5}}right):frac{1}{sqrt{5}-sqrt{2}}
Đọc tiếp
Thực hện phép tính
a, \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
b,\(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
c, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
d, \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}\)
b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)
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Bài 1:
a, \(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-5\sqrt[]{3}\)
b, \(\sqrt[3]{125}+\sqrt[3]{-343}-2\sqrt[3]{64}+\frac{1}{3}\sqrt[3]{216}\)
a, = \(3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)
= \(12\sqrt{2}-2\sqrt{3}\)
b, = 5 - 7 - 8 + 2
= - 8
Bài 1:
a, \(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-\sqrt{75}\)
b, \(\sqrt[3]{125}+\sqrt[3]{-343}-2\sqrt[3]{64}+\frac{1}{3}\sqrt[3]{126}\)
Có
\(\frac{1}{2}\sqrt{72}+\frac{3}{4}\sqrt{48}+\sqrt{162}-\sqrt{75}=3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}=12\sqrt{2}-2\sqrt{3}\)
\(\sqrt[3]{125}+\sqrt[3]{-343}-2\sqrt[3]{64}+\frac{1}{3}\sqrt[3]{126}=5-7-8+\frac{1}{3}\sqrt[3]{126}=\frac{1}{3}\sqrt[3]{126}-10\)