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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)
\(\frac{-7}{6}=\frac{x}{18}=\frac{-98}{y}=\frac{-14}{z}=\frac{t}{102}=\frac{u}{-18}\)
\(\text{+) }\frac{-7}{6}=\frac{x}{18}\)
\(\Rightarrow\left(-7\right).18=6.x\)
\(\Rightarrow-126=6.x\)
\(\Rightarrow x=-21\)
\(\text{+) }\frac{-7}{6}=\frac{-98}{y}\)
\(\Rightarrow\left(-7\right).y=6.\left(-98\right)\)
\(\Rightarrow y=\frac{6.\left(-98\right)}{-7}\)
\(\Rightarrow y=-84\)
\(\text{+) }\frac{-7}{6}=\frac{-14}{z}\)
\(\Rightarrow\left(-7\right).z=\left(-14\right).6\)
\(\Rightarrow\left(-7\right).z=-84\)
\(\Rightarrow z=12\)
\(\frac{-7}{6}=\frac{t}{102}\)
\(\Rightarrow\left(-7\right).102=t.6\)
\(\Rightarrow t=\frac{\left(-7\right).102}{6}=\frac{\left(-7\right).17.6}{6}\)
\(\Rightarrow t=-119\)
\(\frac{-7}{6}=\frac{u}{-18}\)
\(\Rightarrow\frac{-7}{6}=\frac{-u}{18}\)
\(\Rightarrow\left(-7\right).18=\left(-u\right).6\)
\(\Rightarrow-126=\left(-u\right).6\)
\(\Rightarrow-u=-21\)
\(\Rightarrow u=21\)
`-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:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
\(\dfrac{x}{18}=\dfrac{4}{3}\Rightarrow x=\dfrac{18.4}{3}=24\\ \dfrac{20}{y}=\dfrac{4}{3}\Rightarrow y=\dfrac{20.3}{4}=15\\ \dfrac{z}{21}=\dfrac{4}{3}\Rightarrow z=\dfrac{21.4}{3}=28\)
Ta có:
\(\dfrac{x}{18}\) = \(\dfrac{4}{3}\)
⇒ x = \(\dfrac{4}{3}\) . 18
⇒ x = 24
\(\dfrac{20}{y}\) = \(\dfrac{4}{3}\)
⇒ y = 20 : \(\dfrac{4}{3}\)
⇒ y = 15
\(\dfrac{z}{21}\) = \(\dfrac{4}{3}\)
⇒ z = \(\dfrac{4}{3}\) . 21
⇒ z = 28
⇒ x + y + z = 24 + 15 + 28 = 67
Vậy x + y + z = 67
Chỉ có 3 trường hợp
TH1 :x=2 ;y=4 ;z=2 ...
TH2:x=4:y=2;z=4...
TH3 : bằng nhau
Loại trường hợp 1 và 2 vì u mũ x không bằng các số khác
trường hợp 3 chọn
Tự trình bày mà nạp cho thầy Huệ
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