Question

bài 1 .Rút gọn

B= \(\sqrt{\dfrac{a}{6}}\) +\(\sqrt{\dfrac{2a}{3}}\) +\(\sqrt{\dfrac{3a}{2}}\) (với a\(\ge\) 0)

E =|x-2| +\(\dfrac{\sqrt{x^2}}{x}\)với x <0

I=|x-\(\sqrt{\left(x-1\right)^2}\)| -2x(x<0)

Answer 1

B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)

=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)

=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)

=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)

Answer 2

b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)

\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)

e: \(=2-x-x=2-2x\)

i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)

\(=\left|2x-1\right|-2x\)

=1-2x-2x=1-4x