Tìm x, y, z biết x-1/2 = y-2/3 = z-3/4 và x + y + z = 49

Tìm các số x, y, z biết:

a) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) và \(x+y+z=49\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{x-1+y-2+z-3}{2+3+4}=\dfrac{49-6}{9}=\dfrac{43}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{43}{9}\\\dfrac{y-2}{3}=\dfrac{43}{9}\\\dfrac{z-3}{4}=\dfrac{43}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=\dfrac{86}{9}\\y-2=\dfrac{43}{3}\\z-3=\dfrac{172}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{95}{9}\\y=\dfrac{49}{3}\\z=\dfrac{199}{9}\end{matrix}\right.\)

Vậy \(x=\dfrac{95}{9};y=\dfrac{49}{3};z=\dfrac{199}{9}\)

b) \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\) và \(x+y+z=49\)

Đặt \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}=k\left(k\ne0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}2x=3k\\3y=4k\\4z=5k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3k}{2}\\y=\dfrac{4k}{3}\\z=\dfrac{5k}{4}\end{matrix}\right.\)

Theo giả thiết ta có: \(x+y+z=49\)

\(\Leftrightarrow\dfrac{3k}{2}+\dfrac{4k}{3}+\dfrac{5k}{4}=49\)

\(\Leftrightarrow\dfrac{18k+16k+15k}{12}=\dfrac{588}{12}\)

\(\Leftrightarrow18k+16k+15k=588\)

\(\Leftrightarrow49k=588\)

\(\Leftrightarrow k=12\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3.12}{2}=18\\y=\dfrac{4.12}{3}=16\\z=\dfrac{5.12}{4}=15\end{matrix}\right.\)

Vậy \(x=18;y=16;z=15\)