Question

[Mai Vũ Gia Đạt]

giúp mình với, mình cảm ơn

Answer 1

[Quỳnh Anh]

Trả lời:

\(a,\sqrt{\left(11-6\sqrt{2}\right)^2}+\sqrt{\left(11+6\sqrt{2}\right)^2}\)

\(=\left|11-6\sqrt{2}\right|+\left|11+6\sqrt{2}\right|\)

\(=11-6\sqrt{2}+11+6\sqrt{2}\)

\(=22\)

b, \(\sqrt{\left(10-4\sqrt{6}\right)^2}-\sqrt{\left(10+4\sqrt{6}\right)^2}\)

\(=\left|10-4\sqrt{6}\right|-\left|10+4\sqrt{6}\right|\)

\(=10-4\sqrt{6}-\left(10+4\sqrt{6}\right)\)

\(=10-4\sqrt{6}-10-4\sqrt{6}\)

\(=-8\sqrt{6}\)

c, \(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)

\(=\left|4-\sqrt{5}\right|+\left|1-\sqrt{5}\right|\)

\(=4-\sqrt{5}+\sqrt{5}-1\)

\(=3\)

d, \(\sqrt{\left(7+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\left|7+\sqrt{2}\right|-\left|1-\sqrt{2}\right|\)

\(=7+\sqrt{2}-\left(\sqrt{2}-1\right)\)

\(=7+\sqrt{2}-\sqrt{2}+1\)

\(=8\)

Answer 2

[Quỳnh Anh]

Trả lời:

Bài 2:

a, \(5\sqrt{25a^2}-25a\) với \(a\le0\)

\(=5\sqrt{\left(5a\right)^2}-25a\)

\(=5.\left|5a\right|-25a\)

\(=5.\left(-5a\right)-25a\) (vì \(a\le0\))

\(=-25a-25a=-50a\)

b, \(\sqrt{49a^2}+3a\) với \(a\ge0\) 

\(=\sqrt{\left(7a\right)^2}+3a\)

\(=\left|7a\right|+3a\)

\(=7a+3a\) (vì \(a\ge0\))

\(=10a\) 

c, \(\sqrt{16a^4}+6a^2\)

\(=\sqrt{\left(4a^2\right)^2}+6a^2\)

\(=\left|4a^2\right|+6a^2\)

\(=4a^2+6a^2=10a^2\)

d, \(3\sqrt{9a^6}-6a^3\) với \(a\le0\)

\(=3\sqrt{\left(3a^3\right)^2}-6a^3\)

\(=3.\left|3a^3\right|-6a^3\)

\(=3.\left(-3a^3\right)-6a^3\) (vì \(a\le0\))

\(=-9a^3-6a^3=-15a^3\)