Question

[Duong Thi Nhuong]

Cho x , y , z TLT vs 2 , 3 , 4 ; x , t TLN  vs \(\frac{1}{3}\) , -2 và x + y + z - 2t = 4. Tính \(\frac{x}{2}+\frac{y}{3}-z+t\)

Answer 1

[Nguyễn Lê Phước Thịnh]

Theo đề, ta có:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{1}{3}x=-2t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\\\dfrac{x}{-2}=\dfrac{t}{\dfrac{1}{3}}\end{matrix}\right.\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{-\dfrac{1}{3}}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{t}{-\dfrac{1}{3}}=\dfrac{x+y+z-2t}{2+3+4-2\cdot\dfrac{-1}{3}}=\dfrac{4}{\dfrac{29}{3}}=\dfrac{12}{29}\)

Do đó: x=24/29; y=36/29; z=48/29; t=-4/29

\(\dfrac{x}{2}+\dfrac{y}{3}-z+t=\dfrac{12}{29}+\dfrac{12}{29}-\dfrac{48}{29}+\dfrac{-4}{29}=-\dfrac{28}{29}\)