\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\)

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

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

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x-y+z}{4}=\dfrac{x+2y-z}{5}\Leftrightarrow A=\dfrac{4}{5}\)

Bài làm:

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=K\hept{\begin{cases}x=3K\\y=4K\\z=5K\end{cases}}\)

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

\(\Rightarrow\frac{3K+4K-5K}{3K+2\left(4K\right)-5K}\)

\(\Rightarrow\frac{3K+4K-5K}{3K+8K-5K}\)

\(\Rightarrow\frac{K\left(3+4-5\right)}{K\left(3+8-5\right)}\)

\(\Rightarrow\frac{3+4-5}{3+8-5}=\frac{2}{6}\)

Vậy biểu thức B = \(\frac{2}{6}\)

khó quá ?????????????????????????????????????????????????????

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\left(k\ne0\right)\)\(\Rightarrow\begin{cases}x=3k\\y=4k\\z=5k\end{cases}\)

Ta có: \(b=\frac{x+y-z}{x+2y-z}=\frac{3k+4k-5k}{3k+2.4k-5k}=\frac{2k}{3k+8k-5k}=\frac{2k}{6k}=\frac{1}{3}\)

Giải:
Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\)

\(\Rightarrow x=3k,y=4k,z=5k\)

Ta có:
\(B=\frac{x+y-z}{x+2y-z}=\frac{3k+4k-5k}{3k+8k-5k}=\frac{\left(3+4-5\right)k}{\left(3+8-5\right)k}=\frac{2k}{6k}=\frac{1}{3}\)

Vậy \(B=\frac{1}{3}\)

X/3=Y/4=Z/5 -> X=3 , Y=4 , Z=5 -> 3+4-5/3+(2.4)-5=1/4

1/4 nha bn

a: \(A=31x^2y^3-2xy^3+\dfrac{1}{4}x^2y^2+2\)

\(B=2xy^3+\dfrac{3}{4}x^2y^2-31x^2y^3-x^2-5\)

P=\(A+B=x^2y^2-x^2-3\)

\(A-B=62x^2y^3-4xy^3-\dfrac{1}{2}x^2y^2+x^2+7\)

b: Khi x=6 và y=-1/3 thì \(P=\left(6\cdot\dfrac{-1}{3}\right)^2-6^2-3=4-36-3=1-36=-35\)

a) Ta có: \(A=\left(-\dfrac{3}{4}x^2y^5z^3\right)\cdot\left(\dfrac{5}{3}x^3y^4z^2\right)\)

\(=\left(\dfrac{-3}{4}\cdot\dfrac{5}{3}\right)\cdot\left(x^2\cdot x^3\right)\cdot\left(y^5\cdot y^4\right)\cdot\left(z^3\cdot z^2\right)\)

\(=\dfrac{-5}{4}x^5y^9z^5\)