Ta có : \(x+y=xy=\dfrac{x}{y}\)
\(=>x=xy-y=y.\left(x-1\right)\)
\(=>x:y=y\left(x-1\right):y\)
\(=>x:y=x-1\)
Mà : x+y = x:y ( gt)
\(x+y-\left(x-1\right)=0.x+y-x+1\)
\(=>=0.y+1=0\)
=>\(\left\{{}\begin{matrix}y=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left(\dfrac{1}{2};-1\right)\)
Cho \(x;y;z\ne0\) và x - y - z = 0. Tính giá trị biểu thức:
\(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
Tìm \(x;y\) biết:
a) \(5^{2x-1}=5^{2x-3}+125.4\)
b) \(x-y=xy=\dfrac{x}{y}\left(y\ne0\right)\)
Cho a,b,c \(\ne0\), đôi một khác nhau thoả mãn:
a(y-z)=b(x-z)=c(x+y)
CMR: \(\dfrac{y+z}{a\left(b+c\right)}+\dfrac{z+x}{b\left(a-c\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
Tìm x, y, z, t biết \(\dfrac{1}{4x}=\dfrac{1}{2y}=\dfrac{3}{4z}=\dfrac{1}{t}\) và \(x+y+z+t=10\) \(\left(x,y,z,t\ne0\right)\)
Tìm x, y biết :
\(\left|x+3\right|+\left|x-1\right|=\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\)
Tìm x, y biết :
\(\left(x+y-2\right)^2+7=\dfrac{14}{\left|y-1\right|+\left|y-3\right|}\)
Tìm x;y biết :
\(\dfrac{6}{\left(x-1\right)^2+2}=\left|y-1\right|+\left|y-2\right|+\left|y-3\right|+1\)
Tìm x, y, zϵ R biết: \(\left(4x^2-4x+1\right)^{2022}+\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
Bài 1 . Tìm x thỏa mãn biết : \(\left|x+1\right|+\left|x-1\right|+4x^2=2+4x^2\)
Bài 2 . Tìm x , y biết
a ) \(\left|2x+y+1\right|^{2015}+\left(x-1\right)^{2016}\)
b ) \(\frac{1}{x}+\frac{1}{y}+\frac{2}{xy}=1\left(x,y\in Z,x\ne,y\ne0\right)\)
c ) \(3x-\left|2x+1\right|=2\)
