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The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
We know that we can't have a two digit factor in one of our three digit numbers(it can't fill in one digit), so we are left with:
1,2,3,4,6,8
Now, we can just bash out unique cases:
138
146
226
234
138 = 3! = 6 ways
146 = 3! = 6 ways
226 = 3!/2! = 3 ways
234 = 3! = 6 ways
add all these up, and we get:
6+6+3+6 = 21
a)\(\in\)
b)\(\notin\)
c)\(\subset\)
d)\(\in\)
e)\(\in\)
g)\(\notin\)
\(\frac{P}{Q}=\frac{2}{3}=\frac{100}{150}=\frac{P+100}{Q+150}\Rightarrow P+100=\frac{2}{3}\left(Q+150\right)\)
\(\frac{P+100}{Q+200}=\frac{3}{4}\Rightarrow P+100=\frac{3}{4}\left(Q+200\right)\)
\(\frac{2}{3}\left(Q+150\right)=\frac{3}{4}\left(Q+200\right)\Leftrightarrow Q=-600\)
\(\Rightarrow P=-400\).
e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
Call number to find is xy
10x + y = "a two digit number
if you reverse the digits of your number, the result is a number 20% larger than your number.
10y + x = 1.2(10x+y)
10y + x = 12x + 1.2y
10y - 1.2y = 12x - x
8.8y = 11x
8y = x
y = 1.25x
the only single digit integer that satisfies this
x = 4, y = 5
45 is the two digit number
bạn dịch mk đc ko :)))