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a) 18.17−3.6.7
= 18.17 - 18 . 7
= 18 ( 17 - 7 )
= 18 . 10 = 180
b) 54−6.(17+9)
= 54 - 102 - 54
= - 102
c) 33.(17−5)−17.(33−5)
= 33. 17 - 33. 5 - 17 .33 + 17 . 5
= 17 . 5 - 33 . 5
= 5 ( 17 - 33 )
= 5. (-16)
= - 80
a) 18.17 - 3.6.7
= 18.17 - 18.7
= 18.(17 - 7 )
= 18 . 10
= 180
b) 54 - 6.(17 + 9)
= 6.9 - 6.17 + 6.9
= (6.9 - 6.9) - 6.17
= 0 - 6.17
= 0 - 102
= -102
c) 33.(17 - 5) - 17.(33 - 5)
= 33.17 - 33.5 - 17 . 33 - 17.5
= 33.(17 - 17) - 5.(33 - 17 )
= 33. 0 - 5.16
= 0 - 80
= -80
a) 2575 + 37 - 2576 - 39
= 2576 + 36 - 2576 - 29
= 36 - 29
= 7
b) 34 + 35 + 36 + 37 - 14 - 15 - 16 - 17
= (34 - 14) + (35 - 15) + (36 - 16) + (37 - 17)
= 20 + 20 + 20 + 20
= 20 . 4
= 80
a) 3784 + 23 – 3785 – 15
= 3784 - 3785 + 23 - 15
= -(3785 - 3784) + 8
= -1 + 8
= 8 - 1
= 7
b) 21 + 22 + 23 + 24 – 11 - 12 - 13 - 14
= (21 -11) + (22 - 12) + (23 - 13) + (24 - 14)
= 10 + 10 + 10 + 10
= 40
Sách Giáo Khoa
Bài giải:
a)
3784 + 23 - 3785 - 15 . Áp dụng tính chất giao hoán ta có:
= (3784 - 3785) + (23 - 15)
= -1 + 8
= 7
b) 21 + 22 + 23 + 24 - 11 - 12 - 13 - 14. Áp dụng các tính chất giao hoán và kết hợp ta có:
= (21 - 11) + (22 - 12) + (23 - 13) + (24 - 14).
= 10 + 10 + 10 + 10
= 40
Đáp số: a) 7; b) 40.
sua lai
A) \(\left(44.52.60\right):\left(11.13.15\right)\)
\(=\left(4.11.4.13.4.15\right):\left(11.13.15\right)\)
\(=4^3\left(11.13.15\right):\left(11.13.15\right)\)
\(=64\)
B) \(123.456456-456.123123\)
\(=123.456.1001-456.123.1001\)
\(=56088.1001-56088.1001\)
\(=1001.\left(56088-56088\right)\)
\(=1001.0\)
\(=0\)
C) \(\left(98.7676-9898.76\right):\left(2001.2002....2020\right)\)
\(=\left(98.76.101-98.101.76\right):\left(2001.2002....2020\right)\)
\(=0:\left(2001.2002....2020\right)\)
\(=0\)
Bài 1:
Giải:
Ta có: \(a+b=-3\)
\(\Rightarrow a+b+c=-3+c\)
\(\Rightarrow a+\left(-5\right)=-3+c\)
\(\Rightarrow a-c=\left(-3\right)-\left(-5\right)\)
\(\Rightarrow a-c=2\)
Mà \(c+a=-4\)
\(\Rightarrow a=\left(-4+2\right):2=-1\)
\(\Rightarrow c=\left(-4\right)-\left(-1\right)=-3\)
Lại có: \(b+c=-5\)
\(\Rightarrow b+\left(-3\right)=-5\)
\(\Rightarrow b=-2\)
Vậy bộ số \(\left(a;b;c\right)\) là \(\left(-1;-3;-2\right)\)
Bài 2:
\(P=78+\left|78-129\right|+\left(-29\right)\)
\(\Rightarrow P=78+\left[-\left(78-129\right)\right]+\left(-29\right)\)
\(\Rightarrow P=78+\left(-78\right)+129+\left(-29\right)\)
\(\Rightarrow P=100\)
a, \(A=\frac{2012\cdot2011-1}{2010\cdot2012+2011}=\frac{2012\cdot\left(2010+1\right)-1}{2010\cdot2012+\left(2012-1\right)}=\frac{2012\cdot2010+2012-1}{2012\cdot2010+2012-1}=1\)
b, 10,11 + 11,12 + 12,13 + .... + 97,98 + 98,99 + 99,100
= ( 10 + 11 + 12 + .... + 97 + 98 + 99 ) + ( 0,10 + 0,11 + 0,12 + 0,13 + ... + 0,98 + 0,99 )
= { ( 10 + 99 ) . [ ( 99 - 10 ) : 1 + 1 ] ] : 2 } + { ( 0,10 + 0,99 ) . [ ( 0,99 - 0,10 ) : 0,01 + 1 ] : 2 }
= ( 99 . 90 : 2 ) + ( 1,09 . 90 : 2 )
= 4455 + 49,05
= 4504,05
Bài làm:
Ta có:
\(B=-66\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(B=\left(-66\right).\frac{1}{2}+66.\frac{1}{3}-66.\frac{1}{11}-124.\left(37+63\right)\)
\(B=-33+22-6-124.100\)
\(B=17-12400\)
\(B=-12383\)
\(B=-66.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
\(=-66.\frac{1}{2}-\left(-66\right).\frac{1}{3}+\left(-66\right).\frac{1}{11}+\left(-124\right).37+63.\left(-124\right)\)
\(=-33+22-6+\left(-124\right).\left(37+63\right)\)
\(=-11-6+\left(-124\right).100\)
\(=-17-12400\)
\(=-12417\)
b: \(=\dfrac{1}{2}-\left(\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{2016}=\dfrac{1}{2016}\)
\(\frac{423134.846267-423133}{423133.846267+423134}=\frac{423133.846267+846247-423133}{423133.846267+423134}\)
\(=\frac{423133.846267+423134}{423133.846267+423134}=1\)
\(a.\) \(50\cdot374\cdot2=\left(50\cdot2\right)\cdot347=100\cdot347=34700\)
\(b.\)\(36\cdot97+97\cdot64=97\cdot\left(36+64\right)=97\cdot100=9700\)
\(c.157\cdot289-289\cdot57=289\cdot\left(157-57\right)=289\cdot100=28900\)