\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6x}{24}\\ \Rightarrow1+4y=1+6x\\ \Rightarrow4y=6x\Rightarrow x=\dfrac{2}{3}y\\ \dfrac{1+2y}{18}=\dfrac{1+4y}{24}\Rightarrow24+48y=18+72y\\ \Rightarrow24y=6\Rightarrow y=\dfrac{1}{4}\Rightarrow x=\dfrac{2}{3}\cdot\dfrac{1}{4}=\dfrac{1}{6}\)

\(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6x}{24}\)
\(1+4y=1+6x\)
\(4y=6x\)
\(\dfrac{x}{y}=\dfrac{4}{6}\)

\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)

\(=2x^2-2xy-y^2+2xy\)

\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)

\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)

\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(=5x^2-20xy-4y^2+20xy\)

\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)

\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)

\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)

\(=-xy\left(x+1\right)\)

\(=\dfrac{1}{2}.100\left(100+1\right)=50.101=5050\)

cái đoạn\(-xy\left(x+1\right)\)đổi x+1 thành y+1 nha mik đánh nhầm

\(\dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{40}{24}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)

\(\dfrac{5}{3}+\dfrac{3}{x}=\dfrac{49}{24}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)

Bài 1: Rút gọn biểu thức

a) Ta có: \(x\left(x-y\right)+y\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y\right)\)

\(=x^2-y^2\)

b) Ta có: \(7x\left(4y-x\right)+4y\left(y-7x\right)-\left(4y^2-7x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)

\(=-7x^2+7x\)

a, Ta có : \(x\left(x-y\right)+y\left(x-y\right)\)

\(=\left(x+y\right)\left(x-y\right)=x^2-y^2\)

b, Ta có : \(7x\left(4y-x\right)+4y\left(y-7x\right)-\left(4y^2-7x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)

\(=7x-7x^2=7x\left(1-x\right)\)