Question

[Alayna]

Rút gọn :

e) \(\left(x-4\right)\sqrt{16-8x+x^2}\) với x \(\ge\) 4

f) \(\left(2x-5\right)\sqrt{\dfrac{2}{\left(2x-5\right)^2}}\) với x \(\ne\) \(\dfrac{5}{2}\)

g) \(\sqrt{x-4\sqrt{x-4}}\) với x \(\ge\) 4

Answer 1

[Aki Tsuki]

e/ \(\left(x-4\right)\sqrt{16-8x+x^2}=\left(x-4\right)\sqrt{\left(x-4\right)^2}=\left(x-4\right)\left(x-4\right)=\left(x-4\right)^2\)

f/ \(\left(2x-5\right)\sqrt{\dfrac{2}{\left(2x-5\right)^2}}=\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}\)

+) với \(x>\dfrac{5}{2}\) có: \(\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}=\dfrac{2x-5}{2x-5}\cdot\sqrt{2}=\sqrt{2}\)

+) với \(x< \dfrac{5}{2}\) có:

\(\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}=\dfrac{2x-5}{-\left(2x-5\right)}\cdot\sqrt{2}=-1\cdot\sqrt{2}=-\sqrt{2}\)

g/ \(\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4-2\cdot2\cdot\sqrt{2-4}+4}=\sqrt{\left(\sqrt{x-4}+2\right)^2}=\sqrt{x-4}+2\)