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c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
a) 2011 + 5 . [300- ( 18- 8)2]
= 2011 + 5. ( 300 - 102)
= 2011 + 5. 200
= 3011
b) Số số hạng trong tổng trên là:
( 99 - 1) ; 2 = 1 = 50 (số)
99 + 97 = … = 3 + 1
= ( 99 + 1) . 50 : 2
= 2500
c) Số số hạng trong tổngtrên là:
( 100 - 1) : 3 + 1 = 34 ( số)
100 + 97 + 94 + …+ 4 = 1
= ( 100 + 1) . 34 : 2=1717
d) 99 - 97 + 95 - 93 + … + 3 - 1
= 2 + 2 + 2 + … + 2
= 2. 25
= 50
e) 100 - 97 + 94 - …+ 4 - 1
= 3 + 3 + 3 + … + 3
= 3. 17
= 51
`#3107.101107`
\(\left[2^3\cdot97+8\cdot75-\left(-47\right)\cdot2^3\right]\div10^2\)
\(=\left(2^3\cdot97+2^3\cdot75+47\cdot2^3\right)\div10^2\)
\(=\left[2^3\cdot\left(97+75-47\right)\right]\div10^2\)
\(=\left(2^3\cdot125\right)\div10^2\)
\(=\left(2^3\cdot5^3\right)\div10^2\)
\(=\left(5\cdot2\right)^3\div10^2\)
\(=10^3\div10^2\)
\(=10\)
\(\left(2^3.97+8.75-47.2^3\right):10^2\\ =\left[\left(97-47+75\right).2^3\right]:100\\ =\left[125.8\right]:100\\ =1000:100=10\)
a: \(61\cdot45+61\cdot23-68\cdot51\)
\(=61\left(45+23\right)-68\cdot51\)
\(=68\cdot61-68\cdot51\)
\(=68\left(61-51\right)=68\cdot10=680\)
b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)
\(=75-75+4\cdot8\)
\(=4\cdot8=32\)
c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)
\(=\dfrac{36}{20-30+4^2}\)
\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)
d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)
a) ta có:99 - 97 + 95 - 93 + 91 - 89 + ... + 7 - 5 + 3 - 1
= (99 - 97) + (95 - 93) + ... + (7 - 5) + (3 - 1)
= 25.2=50
b)(125.37.32).4
=(125.4).(37.32)
=500.1184
=100.5.1184
=100.5920
=592000
c)A=5+8+11+14+...+302
Số số hạng của A là
(302-5):3+1=100(số)
Tổng A= (302+5).100:2=15350
b)(125.37.32):4
=125.37.32:4
=125.37.8
=125.8.37
=1000.37
=37000
Sorry bạn mk nhìn nhầm dấu
thực hiện phép tính sau một cách hợp lí:
C=(1+2+3)+(3+4+5)+(5+6+7)+..........+(97+98+99)+(99+100+101)
Ko ghi đề
C = (3.2)+(3.4)+(3.6)+...+(3.98)+(3.100)
C = 3 . (2+4+6+...+100)
C = 3 . 2550
C = 7650
Ko theo công thức nha @@
Nhớ k mk nha ^^
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
Giải:
1) (-8/13:3/7+-5/13:3/7).(-4)3.|-3|/7
=[7/3.(-8/13+-5/13)].-192/7
=[7/3.(-1)].-192/7
=-7/3.-192/7
=64
2) 75%-(5/2+5/3)+(-1/2)2
=3/4-25/6+1/4
=(3/4+1/4)-25/6
=1-25/6
=-19/6
Chúc bạn học tốt!
1) \(\left(\dfrac{-8}{13}:\dfrac{3}{7}+\dfrac{-5}{13}:\dfrac{3}{7}\right).\dfrac{\left(-4\right).|-3|}{7}\)
= \(\left[\left(\dfrac{-8}{13}+\dfrac{-5}{13}\right):\dfrac{3}{7}\right].\dfrac{-64.3}{7}\)
= \(\left[-1:\dfrac{3}{7}\right].\dfrac{-192}{7}\)
= \(\dfrac{-7}{3}.\dfrac{-192}{7}\)
= \(64\)
2) \(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^2\)
= \(\dfrac{3}{4}-\dfrac{25}{6}+\dfrac{1}{4}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)-\dfrac{25}{6}\)
= \(1-\dfrac{25}{6}\)
= \(\dfrac{-19}{6}\)
Chúc bạn học tốt !
-7/23*75/97+75/97-11/23*75/97
=75/97*(-7/23+11/23)
=75/97*4/23
=300/2231
(3/8+ -1/4+5/12):2/3
=(9/24+ -6/24+10/24):2/3
=13/24/2/3
=13/16
ai h cho minh minh h lai cho