Question

Rút gọn:\(\frac{\sqrt{9+12a+4a^2}}{\sqrt{b^2}}\)với 0>a>=\(\frac{-3}{2}\), b<0

 

Answer 1

\(\frac{\sqrt{9+12a+4a^2}}{\sqrt{b^2}}\)

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

\(=\frac{2a+3}{-b}\)( theo điều kiện )

Answer 2

bố mày đéo biết

Answer 3

\(\frac{\sqrt{9+12a+4a^2}}{\sqrt{b^2}}\)

\(=\frac{\sqrt{3^2+2.3.2a+\left(2a\right)^2}}{\sqrt{b^2}}\)

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

\(=\frac{|3+2a|}{|b|}\)

Vì \(b< 0\Rightarrow|b|=-b\)

Vì  \(a\ge-1.5\)  nên \(3+2a\ge0\)

Vậy : | 3 + 2a | = 3 + 2a