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BH
12 tháng 8 2019
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
9 tháng 10 2017
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) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
OM
7 tháng 8 2018
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
15 tháng 8 2019
1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)
\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)
4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)
4A = \(8\sqrt{2}+1\)
⇒ A = \(\frac{\text{}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)
2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)
4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)
4B = \(12\sqrt{2}+1\)
⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)
3. \(\sqrt{2-\sqrt{3}}\). \(\left(\sqrt{6}+\sqrt{2}\right)\)
= \(\sqrt{2-\sqrt{3}}\). \(\sqrt{2}.\left(\sqrt{3}+1\right)\)
= \(\sqrt{4-2\sqrt{3}}\) . \(\left(\sqrt{3}+1\right)\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}\) . \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}-1\right)\). \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}\right)^2\) - 12
= 3 - 1
= 2
4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)
= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
= \(\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\) . \(\left(\sqrt{7}-\sqrt{3}\right)\)
= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)
= \(\sqrt{7}\) . (7 - 3)
= 4\(\sqrt{7}\)
5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{3+\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)
= \(\left(\sqrt{10}\right)^2-\left(\sqrt{2}\right)^2\)
= 10 - 2
= 8
6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)
= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)
= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30
= 2
7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)
= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)
= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)
= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)
= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)
TICK MÌNH NHA!
24 tháng 7 2019
\(\sqrt{9-4\sqrt{5}}\)
=\(\sqrt{5-4\sqrt{5}+4}\)
=\(\sqrt{\left(\sqrt{5}-2\right)^2}\)
=\(\sqrt{5}-2\)
24 tháng 7 2019
\(\sqrt{16-2\sqrt{55}}\)
=\(\sqrt{11-2\sqrt{11}.\sqrt{5}+5}\)
=\(\sqrt{\left(\sqrt{11}-\sqrt{5}\right)^2}\)
=\(\sqrt{11}-\sqrt{5}\)
3 tháng 7 2018
xin chào em mới học dưới lớp tám thôi khó quá không biết làm
\(\sqrt{15-2.\sqrt{15}+1}+\sqrt{9+2.3.\sqrt{5}+5}=\sqrt{15}-1+3+\sqrt{5}\)
tự lm nốt