\(a,\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\\ b,\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ \Rightarrow x=-2.\left(-2\right)=4;y=-2.5=-10\)
a)
\(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \left|x-\dfrac{1}{2}\right|=\dfrac{1}{3}\\ \left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}\\x=-\dfrac{1}{3}+\dfrac{1}{2}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)
b)
\(\dfrac{x}{-2}=\dfrac{y}{5}\)
mà `x-y=14` nên áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ =>\left\{{}\begin{matrix}x=-2\cdot\left(-2\right)=4\\y=-2\cdot5=-10\end{matrix}\right.\)
a) |x - 1/2| + 1/3 = 2/3
|x - 1/2| = 2/3 - 1/3
|x - 1/2| = 1/3
x - 1/2 = 1/3 và x - 1/2 = -1/3
*) x - 1/2 = 1/3
x = 1/3 + 1/2
x = 5/6
*) x - 1/2 = -1/3
x = -1/3 + 1/2
x = 1/6
Vậy x = 1/6; x = 5/6
b) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
x/(-2) = y/5 = (x - y)/(-2 - 5) = -14/(-7) = 2
*) x/(-2) = 2 ⇒ x = -2.2 = -4
*) y/5 = 2 ⇒ y = 5.2 = 10
Vậy x = -4; y = 10
