Tham khảo!

\(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\dfrac{-7}{6}=\dfrac{x}{18}\Leftrightarrow6.x=\left(-7\right).18\Rightarrow x=\dfrac{\left(-7\right).18}{6}=-21\)

\(\dfrac{-7}{6}=\dfrac{-98}{y}\Leftrightarrow\left(-7\right).y=6.\left(-98\right)\Rightarrow y=\dfrac{6.\left(-98\right)}{-7}=84\)

\(\dfrac{-7}{6}=\dfrac{-14}{z}\Leftrightarrow\left(-7\right).z=6.\left(-14\right)\Rightarrow z=\dfrac{6.\left(-14\right)}{-7}=12\)

\(\dfrac{-7}{6}=\dfrac{t}{102}\Leftrightarrow6.t=\left(-7\right).102\Rightarrow t=\dfrac{\left(-7\right).102}{6}=-119\)

\(\dfrac{-7}{6}=\dfrac{u}{-78}\Leftrightarrow6.u=\left(-7\right).\left(-78\right)\Rightarrow u=\dfrac{\left(-7\right).\left(-78\right)}{6}=91\)

\(\text{Vậy }x=-21;y=84;y=84;z=12;t=-119;u=91\)

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)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

`-7/6=x/18`

`=>-21/18=x/18`

`=>x=-21(TM\ x in Z)`

`-7/6=-98/y`

`=>-98/84=-98/y`

`=>y=84(TM\ y in Z)`

`-7/6=-14/z`

`=>-14/12=-14/z`

`=>z=12(TM\ z in Z)`

`-7/6=t/102`

`=>-119/102=t/102`

`=>t=-119(TM\ t in Z)`

Vậy `(x,y,z,t)=(-21,84,12,-119)`

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)

câu cuối làm tương tự

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)
2) \(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

3) \(\dfrac{x}{-3}=\dfrac{4}{y}\Rightarrow\left\{{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)

Tìm x, y, z biết: x,y,z tỉ lệ nghịch với 3 : 5 : 6 và x +y+ z= 42

A. x= 18; y= 14; z= 10.

B. x = 20; y = 12; z = 10.

C. x= 16; y=14.; z=12.

D. x= 20; y=10 ; z= 12.

Chúc bạn học tốt!

Tìm x, y, z biết: x,y,z tỉ lệ nghịch với 3 : 5 : 6 và x +y+ z= 42

A. x= 18; y= 14; z= 10.

B. x=20; y= 12; z= 10.

C. x= 16; y=14.; z=12.

D. x= 20; y=10 ; z= 12.

B.x=20;y=12;z=10

Nhớ tick cho mình nhé!