Question

[Trần Hà My]

Bài 3 viết các biểu thức sau thành 1 bình phương

a, \(4+2\sqrt{3}\) b, \(7+4\sqrt{3}\)

c, \(9+4\sqrt{5}\) d, \(31+10\sqrt{6}\)

e, \(13+4\sqrt{3}\) g, \(21+12\sqrt{3}\)

h, \(29+12\sqrt{5}\) i, \(49+8\sqrt{3}\)

k, \(14-6\sqrt{3}\) l,\(23-8\sqrt{7}\)

m, \(15-4\sqrt{11}\) n, \(28-10\sqrt{5}\)

o, \(17-12\sqrt{2}\) p, \(43-30\sqrt{2}\)

q,\(51-10\sqrt{2}\) r, \(49-12\sqrt{5}\)

HELP ME!!!!!

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 3:

a) Ta có: \(4+2\sqrt{3}\)

\(=3+2\cdot\sqrt{3}\cdot1+1\)

\(=\left(\sqrt{3}+1\right)^2\)

b) Ta có: \(7+4\sqrt{3}\)

\(=4+2\cdot2\cdot\sqrt{3}+3\)

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

c) Ta có: \(9+4\sqrt{5}\)

\(=5+2\cdot\sqrt{5}\cdot2+4\)

\(=\left(\sqrt{5}+2\right)^2\)

d) Ta có: \(31+10\sqrt{6}\)

\(=25+2\cdot5\cdot\sqrt{6}+6\)

\(=\left(5+\sqrt{6}\right)^2\)

e) Ta có: \(13+4\sqrt{3}\)

\(=12+2\cdot2\sqrt{3}\cdot1+1\)

\(=\left(2\sqrt{3}+1\right)^2\)

g) Ta có: \(21+12\sqrt{3}\)

\(=12+2\cdot2\sqrt{3}\cdot3+9\)

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

h) Ta có: \(29+12\sqrt{5}\)

\(=20+2\cdot2\sqrt{5}\cdot3+3\)

\(=\left(2\sqrt{5}+3\right)^2\)

i) Ta có: \(49+8\sqrt{3}\)

\(=48+2\cdot4\sqrt{3}\cdot1\)

\(=\left(4\sqrt{3}+1\right)^2\)

k) Sửa đề: \(14-6\sqrt{5}\)

Ta có: \(14-6\sqrt{5}\)

\(=9-2\cdot3\cdot\sqrt{5}+5\)

\(=\left(3-\sqrt{5}\right)^2\)

l) Ta có: \(23-8\sqrt{7}\)

\(=16-2\cdot4\cdot\sqrt{7}+7\)

\(=\left(4-\sqrt{7}\right)^2\)

m) Ta có: \(15-4\sqrt{11}\)

\(=11-2\cdot\sqrt{11}\cdot2+4\)

\(=\left(\sqrt{11}-2\right)^2\)

n) Sửa đề: \(28-10\sqrt{3}\)

Ta có: \(28-10\sqrt{3}\)

\(=25-2\cdot5\cdot\sqrt{3}+3\)

\(=\left(5-\sqrt{3}\right)^2\)

o) Ta có: \(17-12\sqrt{2}\)

\(=9-2\cdot3\cdot2\sqrt{2}+8\)

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

p) Ta có: \(43-30\sqrt{2}\)

\(=25-2\cdot5\cdot3\sqrt{2}+18\)

\(=\left(5-3\sqrt{2}\right)^2\)

q) Ta có: \(51-10\sqrt{2}\)

\(=50-2\cdot5\sqrt{2}\cdot1\)

\(=\left(5\sqrt{2}-1\right)^2\)

r) Ta có: \(49-12\sqrt{5}\)

\(=45-2\cdot3\sqrt{5}\cdot2+4\)

\(=\left(3\sqrt{5}-2\right)^2\)