Question

Bài 1:

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

b)\(\sqrt{11-6\sqrt{2}}+3\sqrt{2}-3\)

c) \(\sqrt{25x^2}+3x\)  với x< 0

d) \(\sqrt{x^2-10x+25}-x\) với x≥5
 

Answer 1

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

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

\(=3+\sqrt{2}+3-\sqrt{2}\)

\(=6\)

c) \(\sqrt{25x^2}+3x\)

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

\(=-5x-3x\) (vì `x<0`)

\(=-8x\)

d) \(\sqrt{x^2-10x+25}-x\)

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

\(=x-5-x\) (vì `x≥5`)

\(=-5\)

 

 

Answer 2

b) \(\sqrt{11-6\sqrt{2}}+3\sqrt{2}-3\)

\(=\sqrt{3^2-2.3.\sqrt{2}+2}+3\sqrt{2}-3\)

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

\(=3-\sqrt{2}+3\sqrt{2}-3=2\sqrt{2}\)