Tìm x,y biết:
\(\dfrac{x}{4}=\dfrac{2y+1}{3}=\dfrac{x-2y-1}{y}\)(với \(y\ne0\))
Biết rằng: \(\dfrac{x+3y}{x-2y}=\dfrac{4}{3},\left(x-2y\ne0\right)\). Khi đó \(\dfrac{x}{y}\left(y\ne0\right)\) bằng:
\(\Leftrightarrow3x+9y=4x-8y\)
\(\Leftrightarrow x=17y\)
hay \(\dfrac{x}{y}=\dfrac{17}{1}\)
\(\Leftrightarrow3\left(x+3y\right)=4\left(x-2y\right)\\ \Leftrightarrow3x+9y=4x-8y\\ \Leftrightarrow x=17y\Leftrightarrow\dfrac{x}{y}=17\)
Giúp mình vớiiiii ! Mình cảm ơn !
Tìm x,y biết : \(\dfrac{x}{4}\) = \(\dfrac{2y+1}{3}\) = \(\dfrac{x-2y-1}{y}\) ( với y ≠ 0 )
Bạn tham khảo tại đây:
https://hoc24.vn/cau-hoi/giup-minh-voiiiii-minh-cam-on-tim-xy-biet-dfracx4-dfrac2y13-dfracx-2y-1y-voi-y-0.4107067269450
Cho x+y=1 và \(xy\ne0\). CMR: \(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2.\left(x+y\right)}{x^2y^2+3}=0\)
\(xy\ne0,x,y\ne1\)
\(A=\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x+y\right)}{x^2y^2+3}\)
\(xét:\dfrac{2\left(x+y\right)}{x^2y^2+3}=\dfrac{2}{x^2y^2+3}\left(1\right)\)
\(\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}\left(2\right)\)
\(xét:\) \(x^4-x-y^4+y=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3-1\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)+xy\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(1-3xy+xy-1\right)\)
\(=\left(x-y\right)\left(-2xy\right)=-2xy\left(x-y\right)=2xy\)
\(xét\) \(\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)
\(=x^3y^3-\left(1-3xy\right)+1=x^3y^3+3xy=xy\left(x^2y^2+3\right)\)
\(\Rightarrow\left(2\right)\Leftrightarrow\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)
\(\left(1\right)\left(2\right)\Rightarrow A=\dfrac{2}{x^2y^2+3}-\dfrac{2\left(x-y\right)}{x^2y^2+3}=\dfrac{2-2x+2y}{x^2y^2+3}\ne0\left(đề-sai\right)\)
Cho x+y=1 \(\left(x,y\ne0\right)\)
chứng minh: \(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{z\left(x-y\right)}{x^2y^2+3}\ne0\)
Cho x+y=1 \(\left(x,y\ne0\right)\)
chứng minh: \(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{z\left(x-y\right)}{x^2y^2+3}\ne0\)
Cho x+y=1 \(\left(x,y\ne0\right)\)
chứng minh: \(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{z\left(x-y\right)}{x^2y^2+3}\ne0\)
Tìm x,y,z biết: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) biết x-2y+3z=-10
Tìm x, y, z biết:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và x - 2y + 3z = 14.