Question

\(\frac{x-1}{2}\)=\(\frac{y-2}{3}\)=\(\frac{z-3}{4}\)va 2x+3y-z=95

Answer 1

Áp dụng tc của dãy tỉ số bằng nhau :

\(\frac{2.\left(x-1\right)}{4}=\frac{3.\left(y-2\right)}{9}=\frac{z-3}{4}=\frac{2x+3x-z-2-6+3}{4+9-4}=\frac{90}{9}=10\)

\(=>\hept{\begin{cases}\frac{x-1}{2}=10=>x-1=20=>x=21\\\frac{y-2}{3}=10=>y-2=30=>y=32\\\frac{z-3}{4}=10=>z-3=40=>z=43\end{cases}}\)

Vậy ...

Answer 2

Trả lời:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

Đặt\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

\(\Rightarrow\hept{\begin{cases}x-1=2k\\y-2=3k\\z-3=4k\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}}\)

Mà\(2x+3y-z=95\)

\(\Rightarrow2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=95\)

\(\Leftrightarrow4k+2+9k+6-4k-3=95\)

\(\Leftrightarrow9k+5=95\)

\(\Leftrightarrow9k=90\)

\(\Leftrightarrow k=10\)

\(\Rightarrow\hept{\begin{cases}x=2.10+1=21\\y=3.10+2=32\\z=4.10+3=43\end{cases}}\)(Thỏa mãn)

Vậy\(\hept{\begin{cases}x=21\\y=32\\z=43\end{cases}}\)

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