Question

Tìm x biếi: \(\dfrac{3x-40}{50}+\dfrac{3x-10}{40}+\dfrac{x+30}{20}+\dfrac{x-90}{10}=0\)

Answer 1

\(\dfrac{3x-40}{50}+\dfrac{3x-10+2x+60+4x-360}{40}=0\)

=> \(\dfrac{3x-40}{50}+\dfrac{9x-310}{40}=0\)

=> \(\dfrac{3x-40}{50}=\dfrac{-9x+310}{40}\)

=> \(40\left(3x-40\right)=50\left(-9x+310\right)\)

=> \(120x-1600=-450x+15500\)

=> \(120x+450x=15500+1600\)

Hay \(570x=17100\)

=>x = 30

Hơi dài nhé bạn

Answer 2

\(\dfrac{3x-40}{50}\)+\(\dfrac{3x-10+2x+60+4x-360}{40}\)=0

⇒\(\dfrac{3x-40}{50}\)+\(\dfrac{9x-310}{40}\)=0

⇒\(\dfrac{3x-40}{50}\)=\(\dfrac{9x-310}{40}\)

⇒40(3x -40) = 50(-9x+310)

⇒120x - 1600 = -450x + 15500

⇒120x + 450x = 15500 + 1600

Mặt khác: 570x = 17100

⇒x = 30