Question

1,Chứng minh

a,11+\(6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)

b,\(8-2\sqrt{7}=\left(\sqrt{7}-1\right)^2\)

c,\(\left(5-\sqrt{3}\right)^2=28-10\sqrt{3}\)

d,\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=2\)

Answer 1

1: Chứng minh

a) Ta có: \(VT=11+6\sqrt{2}\)

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

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

b) Ta có: \(VP=\left(\sqrt{7}-1\right)^2\)

\(=7-2\cdot\sqrt{7}\cdot1+1\)

\(=8-2\sqrt{7}=VT\)(đpcm)

c) Ta có: \(VT=\left(5-\sqrt{3}\right)^2\)

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

\(=28-10\sqrt{3}=VP\)(đpcm)

d) Ta có: \(VP=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

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

\(=\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|\)

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

\(=\sqrt{3}+1-\sqrt{3}+1\)

\(=2=VT\)(đpcm)

Answer 2

Làm câu a :D

a, \(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)

Ở vế phải phân tích HĐT số 2, ta có như sau :

⇔ \(11+6\sqrt{2}\)\(=3^2+2.3.\sqrt{2}+2\)

⇔ 11 \(+6\sqrt{2}=3^2+2+2.3.\sqrt{2}\)(đpcm)