Chứng tỏ
\(\sqrt{3}+\sqrt{8}+1< 6\)
Những câu hỏi liên quan
Chứng tỏ P < \(\frac{8}{9}\)
P=\(\frac{\sqrt{3}-\sqrt{1}}{2}+\frac{\sqrt{5}-\sqrt{3}}{4}+\frac{\sqrt{7}-\sqrt{5}}{6}+...+\frac{\sqrt{81}-\sqrt{79}}{80}\)
Chứng tỏ
\(a,\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6
}\)
\(b,\sqrt{2}+\sqrt{8}< \sqrt{3}+3\)
c,\(\sqrt{5}+\sqrt{10}>5,3\)
a, \(\sqrt{21}>\sqrt{20}\)
\(-\sqrt{5}>-\sqrt{6}\)
\(\Rightarrow\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b, \(\sqrt{2}< \sqrt{3}\)
\(\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\sqrt{2}+\sqrt{8}< \sqrt{3}+3\)
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Chứng tỏ rằng:
a)\(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b)\(\sqrt{2}+\sqrt{8}< \sqrt{3}+3\)
c)\(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
Chứng tỏ
a, \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)
b.\(\sqrt{2}+\sqrt{8}< \sqrt{3}+3\)
c,\(\sqrt{5}+\sqrt{10}>5,3\)
a, Vì
\(\sqrt{21}-\sqrt{5}=2346507717\)
\(\sqrt{20}-\sqrt{6}=2022646212\)
b, Vì
\(\sqrt{2}+\sqrt{8}=4242640687\)
\(\sqrt{3}+3=4732050808\)
c, Vì
\(\sqrt{5}+\sqrt{10}=5398345638\)
\(5,3=5,3\)
P/s; Ủa tôi tưởng lớp 8 mới học về Căn thức chứ
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Ta biết căn( \(\sqrt{ }\)) càng lớn thì càng chia ra số nhỏ
=> a >
b<
c>
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Chứng tỏ: \(\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}.9\sqrt{3}-11\sqrt{3}\)
Là một số nguyên.
chứng tỏ \(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)=\sqrt{3}\)
cho P = \(a-\left(\frac{1}{\sqrt{a}-\sqrt{a-1}}-\frac{1}{\sqrt{a}+\sqrt{a-1}}\right);\left(a\ge1\right)\). chứng tỏ \(P\ge0\)
Lời giải:
Yêu cầu 1:
\(\frac{5+3\sqrt{5}}{\sqrt{5}}+\frac{3+\sqrt{3}}{\sqrt{3}+1}-(\sqrt{5}+3)=\frac{\sqrt{5}(\sqrt{5}+3)}{\sqrt{5}}+\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1}-(\sqrt{5}+3)\)
\(=\sqrt{5}+3+\sqrt{3}-(\sqrt{5}+3)=\sqrt{3}\) (đpcm)
---------
Yêu cầu 2:
\(P=a-\frac{\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a-1}}{(\sqrt{a}-\sqrt{a-1})(\sqrt{a}+\sqrt{a-1})}=a-\frac{2\sqrt{a-1}}{a-(a-1)}=a-2\sqrt{a-1}\)
\(=(a-1)-2\sqrt{a-1}+1=(\sqrt{a-1}-1)^2\geq 0\) với mọi $a\geq 1$
Ta có đpcm.
Chứng minh các đẳng thức sau:
a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{2}\)
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HELP ME:
Chứng tỏ: \(\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}.9\sqrt{3}-11\sqrt{3}\)
Là một số nguyên..
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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