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1) Do x chia hết cho 15, x chia hết cho 25
=> x \(\in\)BC ( 15;25 )
Mà \(15=3.5\)
\(25=5^2\)
=> BCNN ( 15,25 ) = \(5^2.3=75\)
=> BC ( 15;25 ) = B ( 75 ) = { 0 ; 75 ; 150 ; 225 ; ...}
Mà 75 < x < 200
=> x = { 75 ; 150 }
2) Do 35 chia hết cho x
42 chia hết cho x
=> x \(\in\)ƯC ( 35;42 )
Mà \(35=5.7\)
\(42=2.3.7\)
=> UCLN ( 35,42 ) = 7
=> UC ( 35;42 ) = Ư ( 7 ) = { - 7 ; -1 ; 1 ; 7 }
Mà x > 1
=> x = { 1 ; 7 }
a) 125 : x + ( 32 + 15 ) = 72
125 : x + 47 = 72
125 : x = 72 - 47
125 : x = 25
x = 125 : 25
x = 5
b) ( 35 + 135 ) - 2 . x = 86
170 - 2 . x = 86
2 . x = 170 - 86
2 . x = 84
x = 84 : 2
x = 42
c) 7 . x - 5 = 65
7 . x = 65 + 5
7 . x = 70
x = 70 : 7
x = 10
A) 125 / x + (32 + 15) = 72
125 / x + 47 = 72
125 / x = 72 - 47
125 / x = 25
x = 125 / 25
x = 5
a) 125 : X + (32 + 15) = 72
125 : X + 47 = 72
X + 47 = 72 x 125
X + 47 = 62270
X = 62270 - 47
X = 62223
b) (35 + 15) - 2 x X = 86
50 - 2 x X = 86
50 - X = 86 + 2
50 - X = 88
X = 88+ 50
X = 138
35 x 34 + 35 x 65 + 65 x 75 + 65 x 45
= ( 35 x 34 + 35 x 65 ) + ( 65 x 75 + 65 x 45 )
= 35 x ( 34 + 65 ) + 65 x ( 75 + 45 )
= 35 x 99 + 65 x 120
= 3465 + 7800
= 11265
65 x y + 35 x y = 6000
( 65 + 35 ) x y = 6000
100 x y = 6000
y = 6000 : 100
y = 60 .
Vậy y = 60.
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
\(\left(x+35\right)+\left(x+65\right)=200\)
\(x+35+x+65=200\)
\(2x+100=200\)
\(2x=200-100\)
\(2x=100\)
\(x=100:2\)
\(x=50\)
(\(x\) + 35) + (\(x\) + 65) = 200
\(x\) + 35 + \(x\) + 65 = 200
(\(x\) + \(x\)) + ( 35 + 65) = 200
2\(x\) + 100 = 200
2\(x\) = 200 - 100
2\(x\) = 100
\(x\) = 100 : 2
\(x\) = 50
Vậy \(x\) = 50