`(24x^5-18x^4+30x^3) \div 6x^3`

`= 24x^5 \div 6x^3 -18x^4 \div 6x^3 + 30x^3 \div 6x^3`

`= 4x^2-3x+5`

Ta có: \(2x^4+11x^3+11x^2-24x-36\)

\(=2x^4+4x^3+7x^3+14x^2-3x^2-6x-18x-36\)

\(=\left(x+2\right)\left(2x^3+7x^2-3x-18\right)\)

\(=\left(x+2\right)\left(2x^3+4x^2+3x^2+6x-9x-18\right)\)

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

\(=\left(x+2\right)^2\cdot\left(2x^2+6x-3x-9\right)\)

\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)\)

Ta có: \(x^5+7x^4+21x^3+47x^2+80x+60\)

\(=x^5+4x^4+4x^3+3x^4+12x^3+12x^2+5x^3+20x^2+20x+15x^2+60x+60\)

\(=\left(x^2+4x+4\right)\left(x^3+3x^2+5x+15\right)\)

\(=\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)\)

Ta có: \(\frac{2x^4+11x^3+11x^2-24x-36}{x^5+7x^4+21x^3+47x^2+80x+60}\)

\(=\frac{\left(x+2\right)^2\cdot\left(x+3\right)\left(2x-3\right)}{\left(x+2\right)^2\cdot\left(x+3\right)\left(x^2+5\right)}=\frac{2x-3}{x^2+5}\)

A= (937.1 - 4.5) - (-4.5 + 37.1) -100

  = (937-20) - 17 -100

  = 917- 17-100

  =900-100=800

nếu câu trả lời đúng thì bạn k giùm mik nhé?!

(2x+2⁴).5³=4.5⁵

=> 2x+2⁴=4.5⁵:5³

=> 2x+16=4.5²

=> 2x+16=4.25

=> 2x+16=100

=> 2x=100-16

=> x=84:2

=> x=42

(2x+24).53=4.55

=> 2x +2^4 = 4.5^5:5^3

=> 2x+16 = 4.5^2 

=> 2x+16 = 4.25

=> 2x+16= 100 

=> 2x= 100-16

x=84 

=> x= 84:2 

x=42

a/|x|-2,5=27,5

=>|x|=27,5+2,5=30

=>x=30 hoặc x=-30

b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)

=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)

=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)

c/(x-1)\(^5\)=-32

=>x-1=-2 vì (-2)\(^5\)=-32

=>x=-2+1=-1

d/\(\dfrac{4}{5}.x+0,5=4.5\)

=>\(\dfrac{4}{5}.x+0,5=20\)

=>\(\dfrac{4}{5}.x=20-0,5=19,5\)

=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)

\(\dfrac{16x^3y^4}{24x^2y^5}=\dfrac{2x}{3y}\)