\(\frac{1}{\sqrt{x}-\sqrt{x-1}}\) + \(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\)
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M A B C 1
xét tam giác MCA và tam giác MAB có C1 = MAB ( 2 góc cùng chắn cung AB )
góc M chung
=> tam giác MCA đồng dạng tam giác MAB (g-g )
=> MA2 = MB.MC
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\(5+2\sqrt{21}=\left(3+2\sqrt{3.7}+7\right)-5\)=\(\left(\sqrt{3}+\sqrt{7}\right)^2-5=\left(\sqrt{3}+\sqrt{7}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{7}+\sqrt{5}\right)\)
Do đó \(\frac{5+2\sqrt{21}}{\sqrt{7}+\sqrt{3}-\sqrt{5}}=\frac{\left(\sqrt{7}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{3}\sqrt{5}\right)}{\sqrt{7}+\sqrt{3}-\sqrt{5}}\)\(=\sqrt{3}+\sqrt{7}+\sqrt{5}\)
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=10( (1-√4)/(1-4) + (√4-√7)/(4-7)+.....+(√97-√100)/(97-100) )
=10 (1-100)/3
=-990/3 = -330
Mik cx l9
k hay ko tùy bn
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\(2016^3-2016=2016.\left(2016^2-1\right)\)
\(=2016.\left(2016-1\right).\left(2016+1\right)\)
\(=2017.2016.2015⋮2017\) ( đpcm )
20163-2016=2016(20162-1)=2016.(2016-1)(2016+1)=2015.2016.2017 chia hết cho 2017
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