\(B=-\frac{1}{3}+\frac{2}{5}-\frac{2}{3}-\frac{3}{5}+\frac{1}{5}\)

\(=\left(-\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{2}{5}-\frac{3}{5}+\frac{1}{5}\right)\)

\(=-\frac{3}{3}+0\)

\(=-1\)

=.= hk tốt!!

B=\(\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)

  =\(\frac{-4}{4\cdot3}+\frac{2\cdot9}{5\cdot9}+\frac{\left(-2\right)\cdot3}{3\cdot3}+\frac{\left(-4\right)\cdot7}{5\cdot7}+\frac{6}{5\cdot6}\)

  =\(\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-4}{5}+\frac{1}{5}\)

  = \(\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{-4}{5}+\frac{2}{5}+\frac{1}{5}\right)\)

  =\(\frac{-3}{3}+\frac{-1}{5}\)

  = \(-1+\frac{-1}{5}\)=\(\frac{-5-1}{5}=\frac{-6}{5}\)

-4/12 + 18/45 - 6/9 - 21/35 + 6/30

= -1/3 + 2/5 - 2/3 - 3/5 + 1/5

= (2/5 - 3/5 + 1/5) - (1/3 + 2/3)

= 0 - 1

= -1

\(=-\frac{1}{3}+\frac{2}{5}-\frac{2}{3}-\frac{3}{5}+\frac{1}{5}\)

\(=\left(-\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{2}{5}-\frac{3}{5}+\frac{1}{5}\right)\)

\(=-1+0\)

\(=-1\)

=.= hk tốt!!

Ta có: a=18+24-36+x

a=6+x

a) Để a chia hết cho 6 thì 6+x chia hết cho 6

Vì 6 chia hết cho 6 nên x chia hết cho 6.

Vậy x=6k

b) Để a ko chia hết cho 6 thì 6+x ko chia hết cho 6

Vì 6 chia hết cho 6 nên x ko chia hết cho 6.

Vậy x ≠ 6k

a: a=18+24-36+x

=>a=x+6

Để a⋮6 thì x+6⋮6

=>x⋮6

b: Để a không chia hết cho 6 thì x+6 không chia hết cho 6

=>x không chia hết cho 6

a) 19.64+79.34=19.2.32+38.2.68                           

                           =38.32+38.68

                           = 38.(32+68)= 38.100

                           = 3800

b)35.12+65.13= 35.12+65.(12+1)

                          =  (35+65).12

                          =  100.12

                          =1200

làm nhầm câu b

Come on viết đề bài cho tử tế đi bạn (:

nhiều thế, ai lm nổi

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

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

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

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

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)

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

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

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

a)Vì 35\(⋮\)(2x-1)

=>2x-1 \(\in\)Ư(35)

=>2x-1\(\in\){1;-1;5;-5;7;-7;35;-35}

=>2x\(\in\){2;0;6;-4;8;-6;36;-34}

=>x\(\in\){1;0;3;-2;4;-3;18;-17}

b)18\(⋮\)(x+1)

=>x+1\(\in\)Ư(18)

=>x+1\(\in\){1;-1;2;-2;3;-3;6;-6;9;-9;18;-18}

=>x\(\in\){0;-2;1;-3;2;-4;5;-7;8;-10;17;-19}

a)7/16

b)49/270

\(A=\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)

\(A=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)

\(A=-1+1+\frac{-1}{5}\)

\(A=\frac{-1}{5}\)

\(B=\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)

\(B=\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-3}{5}+\frac{1}{5}\)

\(B=\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{2}{5}+\frac{-3}{5}+\frac{1}{5}\right)\)

\(B=-1+0\)

\(B=-1\)