a) Ta có: \(\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)

\(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-\sqrt{3}+5\cdot\dfrac{2}{\sqrt{3}}\)

\(=-9\sqrt{3}+\dfrac{10}{\sqrt{3}}\)

\(=\dfrac{-27+10}{\sqrt{3}}=\dfrac{-17\sqrt{3}}{3}\)

a) 66,8 : 89 = 0,7505...

b) 8,568 : 3,6 = 2,38

c) 429 : 572 = 0,75

d) 127 : 25,4 = 0,005

Chúc bạn học tốt.

😁😁😁

\(8,=\left(2x-3\right)\left(2x+3\right)\\ 9,=\left(1-5a^2\right)\left(1+5a^2\right)\)

8) \(-9+4x^2=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

9) \(1-25a^4=1-\left(5a^2\right)^2=\left(1-5a^2\right)\left(1+5a^2\right)\)

\(\left(-2022\right)+\left(-250\right)+\left(275+2022\right)\)

\(=\left(-2022\right)-250+275+2022\)

\(=\left[\left(-2022\right)+2022\right]+\left(275-250\right)\)

\(=0+25=25\)

=- 2022 -250+ 275 + 2022

=-2272+2297

=25

_{$\left(-2022\right)+\left(-250\right)+\left(275+2022\right)$}

_{=\left(-2022\right)-250+275+2022$=\left(-2022\right)-250+275+2022$}

_{=\left[\left(-2022\right)+2022\right]+\left(275-250\right)$=\left[\left(-2022\right)+2022\right]+\left(275-250\right)$}

_{=0+25=25$=0+25=25$}

ai giúp mình với ạ

a: Xét tứ giác ADHE có 

\(\widehat{ADH}=\widehat{AEH}=\widehat{DAE}=90^0\)

Do đó: ADHE là hình chữ nhật

\(=\dfrac{2x}{x-5}+\dfrac{x}{x+5}+\dfrac{-20x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}+\dfrac{-20x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x^2+10x+x^2-5x-20x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{3x^2-15x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{3x}{x+5}\)