Tìm số nguyên x, biết:
a) (-3).x = 36; b) (-100) : (x + 5) = -5.
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`A)2/3=x/60`
`=>40/60=x/60`
`=>x=40`
`B)-1/2=y/18`
`=>-9/18=y/18`
`=>y=-9`
`C)3/x=y/35=-36/84`
Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`
`=>3/x=-3/7`
`=>x=-7`
`y/35=-3/7=-15/35`
`=>y=-15`
`D)7/x=y/27=-42/54`
Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`
`=>7/x=-7/9`
`=>x=-9`
`y/27=-7/9=-21/27`
`=>y=-21`
\(a,\left(x+12\right)\left(x-6\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-12\\x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< -12\\x< 6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\)
\(b,\left(10-x\right)\left(3-x\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x< 0\\3-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x>0\\3-x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>10\\x< 3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< 10\\x>3\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+12>0\\x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+12< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>6\\x< -12\end{matrix}\right.\\ \Rightarrow x\in\left\{...;-15;-14;-13;7;8;9;...\right\}\\ b,\Rightarrow\left(x-10\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>10;x< 3\left(\text{loại}\right)\\3< x< 10\end{matrix}\right.\\ \Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)
a,(-12).x= 60-12
(-12).x=48
x=48:(-12)
x = -4
b,(-5).x + 5 = -24+6
(-5).x + 5 = 30
(-5).x = 30-5
(-5).x = 25
x = 25 : (-5)
x = -5
c,
a: =>3x+3=5x-25
=>-2x=-28
hay x=14
b: =>3x+6=-4x+20
=>7x=14
hay x=2
a: =>y/15=-2/3
hay y=-10
b: 2/x=x/18
nên \(x^2=36\)
hay \(x\in\left\{6;-6\right\}\)
c: x/9=16/x
nên \(x^2=144\)
hay \(x\in\left\{12;-12\right\}\)
a. \(-3x=36\)
\(x=\dfrac{36}{-3}=-12\)
Vậy....
b. \(-100:\left(x+5\right)=-5\)
\(x+5=-100:\left(-5\right)\)
\(x+5=20\)
\(x=20-5=15\)
Vậy....
a) (-3)x=36
nên x=-12
Vậy: x=-12
b) (-100):(x+5)=-5
\(\Leftrightarrow x+5=20\)
hay x=15
Vậy: x=15