Question

Bài 1: Tính

a) \(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)

b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

Bài 2

a) Cho x<y<0. Tính (thu gọn) A=\(\sqrt{x^2}+\sqrt{y^2}+\sqrt{\left(x+y\right)^2}\)

b) Tình a,b,c biết a+b+c= \(2\sqrt{a}+2\sqrt{b-3}+2\sqrt{c}\)

Answer 1

Bài 1:

a)Ta có:\(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}-\sqrt{5-2\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{2}-\sqrt{5}+1=-\sqrt{2}+1\)

b)Ta có:\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}.\sqrt{4-\sqrt{15}}.\left(\sqrt{10}-\sqrt{6}\right)=\sqrt{4+\sqrt{15}}.1.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)