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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)
-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
`-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 1
a) (x + 3)(x + 2) = 0
x + 3 = 0 hoặc x + 2 = 0
*) x + 3 = 0
x = 0 - 3
x = -3 (nhận)
*) x + 2 = 0
x = 0 - 2
x = -2 (nhận)
Vậy x = -3; x = -2
b) (7 - x)³ = -8
(7 - x)³ = (-2)³
7 - x = -2
x = 7 + 2
x = 9 (nhận)
Vậy x = 9
Bài 2:
\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)
\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)
\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)
\(\frac{x}{8}=-\frac{6}{12}\Leftrightarrow x=-4\)
\(\frac{-8}{y^2}=-\frac{6}{12}\Leftrightarrow y^2=16\Leftrightarrow y=4\)
\(\frac{z}{-18}=-\frac{6}{12}\Leftrightarrow z=9\)
Chúc bạn học tốt ^_^
Giải:
Theo đề ra, ta có:
\(\dfrac{x}{8}=-\dfrac{7}{y}=\dfrac{z}{-18}=\dfrac{16}{t}=-\dfrac{6}{12}\)
\(\Leftrightarrow\dfrac{x}{8}=-\dfrac{7}{y}=-\dfrac{z}{18}=\dfrac{16}{t}=-\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=-\dfrac{1}{2}\\-\dfrac{7}{y}=-\dfrac{1}{2}\\-\dfrac{z}{18}=-\dfrac{1}{2}\\\dfrac{16}{t}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}.8\\y=-7:\left(-\dfrac{1}{2}\right)\\z=-\dfrac{1}{2}.\left(-18\right)\\t=16:\left(-\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=14\\z=9\\t=-32\end{matrix}\right.\)
Vậy ...