Câu hỏi của Vũ DIễm Hằng

Question

Bài 9: Tìm x, y, z, t nguyên biết

12/-6=x/5=-y/3=z/-17=-t/-9

Bài 10: Tìm x, y, z, t, u biết:

4/3=12/9=8/x=y/21=40/2=16/t=u/111

Bài 11: Tìm x, y, z, t, u biết:

-7/6=x/18=-98/y=-14/z=t/102=u/-78

Mong mn giải nhanh giúp ạ.

Mk đang cần gấp để nộp cho thầy ạ

Nguyễn Lê Phước Thịnh · Accepted Answer

Bài 9:Ta có: $\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}$$\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\dfrac{-y}{3}=-2\\dfrac{-z}{17}=-2\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\-y=-6\-z=-34\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\y=6\z=34\t=-18\end{matrix}\right.$Vậy: (x,y,z,t)=(-10;6;34;-18)Bài 11:Ta có: $\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}$$\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}$Ta có: $\dfrac{x}{18}=\dfrac{-7}{6}$$\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21$Ta có: $\dfrac{-98}{y}=\dfrac{-7}{6}$$\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84$Ta có: $\dfrac{-14}{z}=\dfrac{-7}{6}$$\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12$Ta có: $\dfrac{u}{-78}=\dfrac{-7}{6}$$\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91$Ta có: $\dfrac{t}{102}=\dfrac{-7}{6}$$\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119$Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)Câu trả lời: Bài 9:Ta có: $\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}$$\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2$$\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\dfrac{-y}{3}=-2\\dfrac{-z}{17}=-2\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\-y=-6\-z=-34\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\y=6\z=34\t=-18\end{matrix}\right.$Vậy: (x,y,z,t)=(-10;6;34;-18)Bài 11:Ta có: $\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}$$\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}$Ta có: $\dfrac{x}{18}=\dfrac{-7}{6}$$\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21$Ta có: $\dfrac{-98}{y}=\dfrac{-7}{6}$$\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84$Ta có: $\dfrac{-14}{z}=\dfrac{-7}{6}$$\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12$Ta có: $\dfrac{u}{-78}=\dfrac{-7}{6}$$\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91$Ta có: $\dfrac{t}{102}=\dfrac{-7}{6}$$\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119$Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)