Tìm các số nguyên x, y, z biết :-4/6=x/15-8/y=z/-12
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-6 /12 = x /8 = -7 /y = z /-18
=>-6*8=12*x
-48=12*x
-48:12=x
=>x=-4
thayx:-6 /12=-4/8=-7/y=z/-18
=>-4*y=8*7
-4*y=56
y=56:(-4)
y=14
=>y=14
thayy:-6/12=-4/8=-7/14=z/18
-4*18=8*z
-72=8*z
-72:8=z
-9=z
=>z=-9
vayx=-4;y=14;z=-9
Ta có: \(\frac{-6}{8}=\frac{x}{16}\Rightarrow x=\frac{16.\left(-6\right)}{8}=-12\)
Thế x = -12 \(\Rightarrow\frac{-12}{16}=\frac{-30}{y}\Rightarrow y=\frac{16.\left(-30\right)}{-12}=40\)
Thế y = 40 \(\Rightarrow\frac{-30}{40}=\frac{z}{-4}\Rightarrow z=\frac{\left(-30\right)\left(-4\right)}{40}=3\)
Vậy x = -12 ; y = 40, z = 3
\(\frac{x}{2}=\frac{8}{y}=\frac{-12}{z}=2\)
1. \(\frac{x}{2}=\frac{2}{1}\)Mà \(\frac{2}{1}=\frac{4}{2}\)\(\Rightarrow x=4\)
Ta có :\(\frac{4}{2}=\frac{8}{y}\)\(\Leftrightarrow\frac{8}{4}=\frac{8}{y}\)\(\Rightarrow y=4\)
Ta lại có : \(\frac{-12}{z}=\frac{2}{1}\)\(\Leftrightarrow\frac{-12}{z}=\frac{-12}{-6}\)\(\Rightarrow z=-6\)
K/l : Vậy \(x=4;y=4;z=-6\)
\(\frac{x}{8}=-\frac{6}{12}\Leftrightarrow x=-4\)
\(\frac{-8}{y^2}=-\frac{6}{12}\Leftrightarrow y^2=16\Leftrightarrow y=4\)
\(\frac{z}{-18}=-\frac{6}{12}\Leftrightarrow z=9\)
Chúc bạn học tốt ^_^
\(\frac{12}{-6}=\frac{x}{5}=\frac{-y}{3}=\frac{z}{-17}=\frac{-t}{-9}\)
\(-6x=12\cdot5=60\Rightarrow x=-10\)
\(-y\cdot\left(-6\right)=12\cdot3=36\Rightarrow y=6\)
\(-6z=-17\cdot12=>z=34\)
\(-t\cdot\left(-6\right)=-9\cdot12=>t=-18\)
1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
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)
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