Chứng minh rằng: \(\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^8>3^6\)
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chứng minh rằng
\(\left[\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right]^8>3^6\)
Chứng minh rằng: \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\frac{-3}{2}\)
có VT \(=\left(\frac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}-\frac{6\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}=\left(\frac{\sqrt{3}}{\sqrt{2}}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{-3\sqrt{3}}{\sqrt{2}}.\frac{1}{\sqrt{6}}=\frac{-3}{2}\)
dpcm
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Ta có: \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
\(=\left\{\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}\right]-\frac{6\sqrt{6}}{3}\right\}\times\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right)\times\frac{1}{\sqrt{6}}\)
\(=\left(-\frac{3\sqrt{6}}{2}\right)\times\frac{1}{\sqrt{6}}\)
\(=\frac{-3}{2}\)(đpcm)
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Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
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Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
Chứng minh
\(\left(\sqrt{4}-\sqrt{3}\right)^2=\sqrt{49}-\sqrt{48}\)
\(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)^2+6\sqrt{6}=9\)
\(\sqrt{8-2\sqrt{15}-\sqrt{8+2\sqrt{15}}}=-2\sqrt{3}\)
+) \(\left(\sqrt{4}-\sqrt{3}\right)^2=4-2\sqrt{4\cdot3}+3=7-2\sqrt{7}=\sqrt{49}-\sqrt{48}\)
+) \(2\sqrt{2}\left(2-3\sqrt{3}\right)+\left(1-2\sqrt{2}\right)^2+6\sqrt{6}\)
\(=4\sqrt{2}-6\sqrt{6}+9-4\sqrt{2}+6\sqrt{6}\)
\(=9\)
+) Sửa : \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5-2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(=-2\sqrt{3}\)
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Chứng minh rằng:
a)\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\) là số nguyên
b)\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
chứng minh \(\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^8>3^6\)
Chứng minh :\(\left(\sqrt[3]{3+2.\sqrt{2}}+\sqrt[3]{3-2.\sqrt{2}}\right)^8>3^6\)
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Chứng minh \(\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^8>6\)