c) 128 : (x - 2012) + 36 = 100
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9900 : 36 - 15 x 11
= 275 - 165
= 110
9700 : 100 + 36 x 12
= 97 + 432
= 529
( 15792 : 336 ) x 5 + 27 x 11
= 47 x 5 + 297
= 235 + 297
= 532
( 160 x 5 - 25 x 4 ) : 4
= ( 800 - 100 ) : 4
= 700 : 4
= 175
1036 + 64 x 52 - 1827
= 1036 + 3328 - 1827
= 4364 - 1827
= 2537
215 . 869 + 215 . 14
= 215 . ( 869 + 14 )
= 215 . 883
= 189945
b) (45 x 46 + 47 x 48) x (45 x 128 - 90 x 64) x (2009 x 2010 + 2011 x 2012)
= (45 x 46 + 47 x 48) x (45 x 2 x 64 - 45 x 2 x 64) x (2009 x 2010 + 2011 x 2012)
= (45 x 46 + 47 x 48) x 0 x (2009 x 2010 + 2011 x 2012) = 0
A=-18-8.9+100:20
A=-18-72+5
A=-85
B= \([\left(-13+3\right)]\) +\([\left(-15+5\right)]\) +\([\left(-17+7\right)]\) +(-9)+11
B=10+10+10+(-9)+11
B=32
C=(2-3)+(4-5)+(6-7)+...+(2012-2013)
C=(-1)+(-1)+(-1)+...+(-1)
Số số hạng là :
(2013-2):1+1
=2012
=> C= (-1).2012
C=-2012
Ta có 46 x 128 + 92 x 36 =46 x 128 +(2x46) x 36
=46 x 128 +46x72
=46x(128+72)= 46x200=9200
\(\text{46 x 128 + 92 x 36}\)
= \(\text{46 . 128 +( 2 . 46 ) . 36}\)
= \(\text{46 . 128 +46. ( 36 . 2 ) }\)
\(\text{= 46 . 128 + 46 . 72 }\)
= \(46\left(72+128\right)\)
= \(46.200\)
= \(9200\)
c, C=|x-1|+|x-2|+...+|x-100|=(|x-1|+|100-x|)+(|x-2|+|99-x|)+...+(|x-50|+|56-x|) \(\ge\) |x-1+100-x|+|x-2+99-x|+...+|x-50+56-x|=99+97+...+1 = 2500
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-1\right)\left(100-x\right)\ge0\\\left(x-2\right)\left(99-x\right)\ge0.....\\\left(x-50\right)\left(56-x\right)\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}1\le x\le100\\2\le x\le99....\\50\le x\le56\end{cases}\Leftrightarrow}50\le x\le56}\)
Vậy MinC = 2500 khi 50 =< x =< 56
a. A=|x-2011|+|x-2012|=|x-2011|+|2012-x| \(\ge\) |x-2011+2012-x| = 1
Dấu "=" xảy ra khi \(\left(x-2011\right)\left(2012-x\right)\ge0\Leftrightarrow2011\le x\le2012\)
Vậy MinA = 1 khi 2011 =< x =< 2012
b, B=|x-2010|+|x-2011|+|x-2012|=(|x-2010|+|2012-x|) + |x-2011|
Ta có: \(\left|x-2010\right|+\left|2012-x\right|\ge\left|x-2010+2012-x\right|=0\)
Mà \(\left|x-2011\right|\ge0\forall x\)
\(\Rightarrow B=\left(\left|x-2010\right|+\left|2012-x\right|\right)+\left|x-2011\right|\ge2+0=2\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-2010\right)\left(2012-x\right)\ge0\\\left|x-2011\right|=0\end{cases}\Rightarrow\hept{\begin{cases}2010\le x\le2012\\x=2011\end{cases}\Rightarrow}x=2011}\)
Vậy MinB = 2 khi x = 2011
Câu c để nghĩ
a) 36 x 25 x 4 = 36 x ( 25 x 4) = 36 x 100 = 3600
18 x 24 : 9= 24 x ( 18:9) = 24 x 2 = 48
41 x 2 x 8 x 5= 41 x 8 x ( 5x2) = 41 x 8 x 10 = 3280
b) 108 x (23 + 7)= 108 x 30 = 3240
215 x 86 + 215 x 14 = 215 x ( 86 + 14) = 215 x 100 = 21500
53 x 128 – 43 x 128= 128 x ( 53 - 43) = 128 x 10 = 1280
a) \(6x-36=5^{12}:5^{10}-2012^0\)
\(6x-36=5^2-2012^0\)
\(6x-36=25-1\)
\(6x-36=24\)
\(6x\) \(=24+36\)
\(6x\) \(=60\)
\(x=60:6\)
\(x=10\)
Vậy \(x=10\)
b) \(2x-125=-65\)
\(2x\) \(=\left(-65\right)+125\)
\(2x\) \(=60\)
\(x=60:2\)
\(x=30\)
Vậy \(x=30\)
c) \(2012-120\left(4x+1\right)=932\)
\(120\left(4x+1\right)=2012-932\)
\(120\left(4x+1\right)=1080\)
\(4x+1=1080:120\)
\(4x+1=9\)
\(4x\) \(=9-1\)
\(4x\) \(=8\)
\(x=8:4\)
\(x=2\)
Vậy \(x=2\)
a) 6x - 36 = 512 : 510 - 20120
6x - 36 = 52 - 20120
6x - 36 = 25 - 1
6x - 36 = 24
6x = 24 + 36
6x = 60
x = 60 : 6
x = 10
Vậy x = 10
b) 2x - 125 = -65
2x = (-65) + 125
2x = 60
x = 60 : 2
x = 30
Vậy x = 30
c) 2012 - 120(4x + 1) = 932
120(4x + 1) = 2012 - 932
120(4x + 1) = 1080
4x + 1 = 1080 : 120
4x + 1 = 9
4x = 9 - 1
4x = 8
x = 8 : 4
x = 2
Vậy x = 2.
Chúc bạn học tốt!
128 : ( x - 2012 ) + 36 = 100
128 : ( x - 2012 ) = 64
x - 2012 = 128 : 64
x - 2012 = 2
x = 2012 + 2
x = 2014