Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-6\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-12\sqrt{6}+3^2}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vi \(\sqrt{6}-3< 0\))
\(=\sqrt{6}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
Báo cáo sai phạm
1) 2√5−√125−√80+√605
=2√5−√52.5−√42.5+√112.5
=2√5−5√5−4√5+11√5
=4√5
2) √15−√216+√33−12√6
=√15−√62.6+√33−12√6
=√15−6√6+√33−12√6
=√(√6)2−6√6+32+√(2√6)2−12√6+32
=√(√6−3)2+√(2√6−3)2
=|√6−3|+|2√6−3|
=3−√6+2√6−3 ( vi √6−3<0)
=√6
5) 2√163 −3√127 −6√475
=24√3 −3.13 −6√223.52
=8√33 −1−6.25 .√13
=8√33 −1−125 .√33
=285 .√33 −1
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)
\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}\)
\(=\dfrac{\sqrt{3}+\sqrt{\dfrac{5}{4}-\dfrac{\sqrt{6}}{2}}}{3}+\dfrac{\sqrt{2}}{6}\)
b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)
c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)
d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)
\(=\dfrac{\sqrt{3\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
\(=\dfrac{\sqrt{3-\sqrt{3}-1}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{3}-1\right)\cdot\left(\sqrt{6}+\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+4\sqrt{3}+2\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)
\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)
\(=\dfrac{\sqrt{1\cdot4}}{2}\)
\(=\dfrac{2}{2}\)
\(=1\)
B2:
3) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2020}-\sqrt{2019}}{2020-2019}\)
\(=\sqrt{2}-1+\sqrt{3}-2+...+\sqrt{2020}-\sqrt{2019}\)
\(=\sqrt{2020}-1\)