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\(\frac{-7}{6}=\frac{x}{18}=\frac{-98}{y}=\frac{-14}{z}=\frac{t}{102}=\frac{u}{-18}\)
\(\text{+) }\frac{-7}{6}=\frac{x}{18}\)
\(\Rightarrow\left(-7\right).18=6.x\)
\(\Rightarrow-126=6.x\)
\(\Rightarrow x=-21\)
\(\text{+) }\frac{-7}{6}=\frac{-98}{y}\)
\(\Rightarrow\left(-7\right).y=6.\left(-98\right)\)
\(\Rightarrow y=\frac{6.\left(-98\right)}{-7}\)
\(\Rightarrow y=-84\)
\(\text{+) }\frac{-7}{6}=\frac{-14}{z}\)
\(\Rightarrow\left(-7\right).z=\left(-14\right).6\)
\(\Rightarrow\left(-7\right).z=-84\)
\(\Rightarrow z=12\)
\(\frac{-7}{6}=\frac{t}{102}\)
\(\Rightarrow\left(-7\right).102=t.6\)
\(\Rightarrow t=\frac{\left(-7\right).102}{6}=\frac{\left(-7\right).17.6}{6}\)
\(\Rightarrow t=-119\)
\(\frac{-7}{6}=\frac{u}{-18}\)
\(\Rightarrow\frac{-7}{6}=\frac{-u}{18}\)
\(\Rightarrow\left(-7\right).18=\left(-u\right).6\)
\(\Rightarrow-126=\left(-u\right).6\)
\(\Rightarrow-u=-21\)
\(\Rightarrow u=21\)
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)
`-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:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
Ta có :
\(\frac{-6}{12}=\frac{x}{8}=\frac{-7}{y}=\frac{z}{-18}=\frac{-1}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=\frac{-1}{2}\Rightarrow x=\left(-4\right)\\\frac{-7}{y}=\frac{-1}{2}\Rightarrow y=14\\\frac{z}{-18}=\frac{-1}{2}\Rightarrow z=9\end{cases}}\)
Vậy ...
1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)
\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)
\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)
\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)
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