Question

Bài 11: Cho x, y, z thõa mãn: \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\) với \(x,y,z\) khác 0. Tính: \(P=\frac{x-y+z}{x+2y-z}\)

Answer 1

Có: \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=\frac{2y}{10}\)

AD TC DTSBN, có:

\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=\frac{2y}{10}=\frac{x-y+z}{2-5+7}=\frac{x+2y-z}{2+10-7}=\frac{x-y+z}{4}=\frac{x+2y-z}{5}\)

\(\Rightarrow\)\(\frac{x-y+z}{4}=\frac{x+2y-z}{5}\Rightarrow\frac{4}{5}=\frac{x-y+z}{x+2y-z}\)

VẬy...

Answer 2

Bài 11:

+ Ta có: \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\left(x,y,z\ne0\right).\)

Đặt \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\\z=7k\end{matrix}\right.\)

+ Lại có: \(P=\frac{x-y+z}{x+2y-z}.\)

Thay \(x=2k;y=5k\) và \(z=7k\) vào P ta được:

\(P=\frac{2k-5k+7k}{2k+2.5k-7k}\)

\(\Rightarrow P=\frac{2k-5k+7k}{2k+10k-7k}\)

\(\Rightarrow P=\frac{\left(2-5+7\right).k}{\left(2+10-7\right).k}\)

\(\Rightarrow P=\frac{4k}{5k}\)

\(\Rightarrow P=\frac{4}{5}.\)

Vậy \(P=\frac{4}{5}.\)

Chúc bạn học tốt!

Answer 3

Sao chép rồi dán lên gg nha

Tham khảo : https://hoc24.vn/hoi-dap/question/119945.html