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b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
B = ( 2 - 4 - 6 + 8 ) + ( 10 - 12 - 14 + 16 ) +......+ ( 98 - 100 - 102 + 104 )
B = 0 + 0 +....+ 0 ( 52 số 0 )
B = 0
A=1+(3-5)+(7-9)+...+(2015-2017)
A=1+(-2)+(-2)+...+(-2)
A=(-2)*1008+1
A=(-2014)+1
A=-2015
Bài 1 : 12 - 12 + 11 + 10 - 9 + 8 - 7 + 5 - 4 + 3 + 2 - 1
= ( 12 - 12 ) + ( 11 - 1 ) + ( 10 - 9 ) + ( 8 - 7 ) + ( 5 - 4 ) + ( 3 + 2 )
= 0 + 10 + 1 + 1 + 1 + 5
= 18
Bài 2 :
3x + 27 = 9
3x = 9 - 27
3x = - 18
x = - 6
2x + 12 = 3( x - 7 )
2x + 12 = 3x - 21
3x - 2x = 12 + 21
x = 33
2x2 - 1 = 49
2x2 = 49 + 1
2x2 = 50
x2 = 50 : 2
x2 = 25
=> x = 5 hoặc x = - 5
- | 9 - x | - 5 = 12
- | 9 - x | = 12 + 5
- | 9 - x | = 17
TH1 : 9 - x >= 0 <=> x <= 9
=> - ( 9 - x ) = 17
=> x = 26 ( loại )
TH2 : 9 - x < 0 <=> x > 9
=> - ( 9 - x ) = -17
=> x = - 8 ( loại )
=> ko có giá trị nào thõa mãn
Bài 3 a,: A = ( - a - b + c ) - ( - a - b - c )
= - a - b + c + a + b + c
= 2c
b, thay c = - 2 vào biểu thức A = 2c
Ta được : A = 2 x ( -2 ) = - 4
a mink la nguoi gui cau hoi minh co mot chu y la ra so tu nhien nha mn
cam on
Bạn ơi, Bạn thiếu 1 dấu ngoặc vuông rồi, cho lại đề bài đi.Thiếu một cái là có thể làm sai đấy.
a, \(A=\dfrac{1}{3}.\dfrac{-6}{-3}.\dfrac{-9}{10}.\dfrac{-13}{36}\)
\(A=\dfrac{1.\left(-6\right).\left(-9\right).\left(-13\right)}{3.13.10.36}\)
\(A=\dfrac{-1}{10.2}\)
\(A=\dfrac{-1}{20}\)
b, \(B=\dfrac{-1}{3}.\dfrac{-15}{17}.\dfrac{34}{45}\)
\(B=\dfrac{\left(-1\right).\left(-15\right).34}{3.17.45}\)
\(B=\dfrac{2}{3.3}\)
\(B=\dfrac{2}{9}\)
c, \(C=\dfrac{1}{3}.\dfrac{4}{5}+\dfrac{1}{3}.\dfrac{6}{5}+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.\left(\dfrac{4}{5}+\dfrac{6}{5}\right)+\dfrac{2}{3}\)
\(C=\dfrac{1}{3}.2+\dfrac{2}{3}\)
\(C=\dfrac{2}{3}+\dfrac{2}{3}\)
\(C=\dfrac{4}{3}\)
d, \(D=\dfrac{-5}{6}.\dfrac{4}{19}+\dfrac{-7}{12}.\dfrac{4}{19}-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(D=\dfrac{4}{19}.\dfrac{-17}{12}-\dfrac{40}{57}\)
\(D=\dfrac{-17}{57}-\dfrac{40}{57}\)
\(D=\dfrac{-57}{57}=-1\)
e, \(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{1}{14}.\dfrac{1}{13}-\dfrac{1}{7}\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{14}.\dfrac{1}{13}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\left(\dfrac{1}{182}+\dfrac{1}{7}\right)\)
\(E=\dfrac{3}{7}.\dfrac{9}{26}-\dfrac{27}{182}\)
\(E=\dfrac{27}{182}-\dfrac{27}{182}\)
\(E=0\)
1.
3.10.15 - 6.5.25
= 30.15 - 30.25
= 30.(15 - 25)
= 30.(-10)
= -300
3 . 10 . 15 - 6 . 5 . 25
= 30 . 15 - 30 . 25
= 30 . (15 - 25)
= 30 . (-10)
= -300
A) \(100+\left(-3\right).\left(-5\right)-\left(-2\right)=100+\left(\left(-3\right).\left(-5\right)\right)+2=100+15+2=117\)
B) \(-2-4-6-8-10-12-14=\left(-2-4\right)+\left(-6-14\right)+\left(-8-12\right)+\left(-10\right)\)
\(=-6+\left(-20\right)+\left(-20\right)+\left(-10\right)=-6-20-20-10=-56\)