Question

Answer 1

7,

a,với y=12 thì \(\dfrac{x}{12}=\dfrac{3}{4}\) nên x=9

b,với x-y=6 thì x=y+6 nên \(\dfrac{y+6}{y}=\dfrac{3}{4}=1+\dfrac{6}{y}\) suy ra y=-24 suy ra x=-18

c,với xy=48 thì \(x=\dfrac{48}{y}\) nên \(\dfrac{48}{y^2}=\dfrac{3}{4}\)suy ra \(y^2=64\Rightarrow\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

 

Answer 2

8,\(\dfrac{x}{2}=\dfrac{y}{3},\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x-y+z}{10-15+21}=\dfrac{39}{4}\)

từ đó tìm dc x,y,z

Answer 3

Bài 9:

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

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

Answer 4

9,\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2y}{10}=\dfrac{x-y+z}{2-5+7}=\dfrac{x+2y-z}{2+10-7}\\ \Rightarrow\dfrac{x-y+z}{x+2y-z}=\dfrac{4}{5}\)

tik mik 3 cái nha

Answer 5

Bài 9:

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

Suy ra: x=2k; y=5k; z=7k

Ta có: \(A=\dfrac{x-y+z}{x+2y-z}\)

\(=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{4k}{5k}=\dfrac{4}{5}\)