C

A

\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+...+\dfrac{9998}{9999}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{9999}\right)\\ =\left(1+1+1+...+1\right)\left(\text{có 50 số 1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\right)\\ =50\cdot1-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\right)\\ =50-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =50-\left(1-\dfrac{1}{101}\right)\\ =50-1+\dfrac{1}{101}\\ =49+\dfrac{1}{101}\\ =\dfrac{4949+1}{101}\\ =\dfrac{4950}{101}\)

M=1−31​+1−151​+1−351​+1−631​+...+1−99991​

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)

\(=3,571\)

M=1−31​+1−151​+1−351​+1−631​+...+1−99991​

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)

a) 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10 + 11

= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10) + 11

= -1 + (-1) + (-1) + (-1) + (-1) + 11

= -1 x 5 + 11

= -5 + 11

= 6

b) 98 + 87 + 76 + ... + 21 - 12 - 23 - 34 - ... - 89

= (98  89) + (87 - 78) + ... + (21 - 12) (có 8 nhóm)

= 9 + 9 + ... + 9 (có 8 số 9)

= 9 x 8

= 72

\(1-2+3-4+.....-10+11\)

\(=\left(1-2\right)+\left(3-4\right)+...\left(9-10\right)+11\)

\(=1+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+11\)

\(=-1\times5+11\)

\(=6\)

\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}\)

\(=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+...+\left(1-\frac{1}{9999}\right)\)

\(=\left(1-\frac{1}{1\cdot3}\right)+\left(1-\frac{1}{3\cdot5}\right)+\left(1-\frac{1}{5\cdot7}\right)+\left(1-\frac{1}{7\cdot9}\right)+...+\left(1-\frac{1}{99\cdot101}\right)\)

\(=\left(1+1+1+1+...+1\right)-\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

Có tất cả : (101 - 3) : 2 + 1 = 50 chữ số 1 => (1 + 1 + 1 + .... + 1) = 1 x 50 = 50 

\(\Rightarrow50-\frac{1}{2}\cdot\left(1-\frac{1}{101}\right)\)

\(=50-\frac{1}{2}\cdot\frac{100}{101}=50-\frac{100}{101}=\frac{4950}{101}\)

Vậy \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+...+\frac{9998}{9999}=\frac{4950}{101}\)

M=1−31​+1−151​+1−351​+1−631​+...+1−99991​

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)

bạn ơi tách ra thừa số chung rồi làm như bình thường nha 

1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)

     =4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)

     = 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)

    =4-\(\left(1-\frac{1}{9}\right)\)

     = 4-\(\frac{8}{9}\)

      = \(\frac{7}{9}\)

M=1−31​+1−151​+1−351​+1−631​+...+1−99991​

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)

\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)

\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)

\(\frac{7}{8}+\frac{18}{10}=\frac{35}{40}+\frac{72}{40}=\frac{107}{40}\)

\(\frac{123}{15}-\frac{85}{13}=\frac{1599}{195}-\frac{1275}{195}=\frac{324}{195}=\frac{108}{65}\)

\(\frac{78}{12}\times\frac{87}{89}=\frac{6786}{1068}=\frac{2262}{356}\)

1) -I-5I

=I - 5 I = 5

2 - ( - 3 )

= - 1 = I - 1 I

= 1

2) ( -87 + 35 ) - ( -34 - 87 )

= -87 + 35 - -34 - 87

= ( 87 + ( -87 ) ) - ( 35 - (-34)

= 0-1

= -1

b) (-168-37)-(-12+137)

= (-168)-37 - (-12)+137

= ((-168)-(-12)) + (137-37)

= (-190)+100

=-90

Số đối của -|-5| là 5

Số đối của 2 - (-3) là -5