Câu hỏi của Tuyet Thanh Tran

Question

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$\dfrac{1}{2}\sqrt{16a^2}-\sqrt{9a^2}$

Tú Nguyễn · Accepted Answer

https://i.imgur.com/oWgIHgn.jpgCâu trả lời: https://i.imgur.com/oWgIHgn.jpg

Nào Ai Biết · Answer

$\dfrac{1}{2}\sqrt{16a^2}-\sqrt{9a^2}$

$=\dfrac{1}{2}\sqrt{\left(4a\right)^2}-\sqrt{\left(3a\right)^2}$

$=\dfrac{1}{2}\left|4a\right|-\left|3a\right|$

$=\dfrac{1}{2}\left|a\right|$Câu trả lời: $\dfrac{1}{2}\sqrt{16a^2}-\sqrt{9a^2}$

$=\dfrac{1}{2}\sqrt{\left(4a\right)^2}-\sqrt{\left(3a\right)^2}$

$=\dfrac{1}{2}\left|4a\right|-\left|3a\right|$

$=\dfrac{1}{2}\left|a\right|$

Nào Ai Biết · Answer

Nhầm ha.

$\dfrac{1}{2}\sqrt{16a^2}-\sqrt{9a^2}$

$=\dfrac{1}{2}.4\left|a\right|-3\left|a\right|$

$=2\left|a\right|-3\left|a\right|$

$=-\left|a\right|$Câu trả lời: Nhầm ha.

$\dfrac{1}{2}\sqrt{16a^2}-\sqrt{9a^2}$

$=\dfrac{1}{2}.4\left|a\right|-3\left|a\right|$

$=2\left|a\right|-3\left|a\right|$

$=-\left|a\right|$

Bài 1: Căn bậc hai