C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

  =\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

  =\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

  =\(\frac{1}{100}-\frac{99}{100}\)

  =\(\frac{-98}{100}=\frac{-49}{50}\)

C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1 
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1) 
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A 
Dễ thấy 1/2.1 = 1/1 - 1/2 
1/3.2 = 1/2 - 1/3 
..................... 
1/99.98 = 1/98 - 1/99 
1/100.99 = 1/99 - 1/100 
=> cộng từng vế với vế ta

\(B=\left(\frac{3}{7}+\frac{-3}{7}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{5}{9}+\frac{-5}{9}\right)+\left(\frac{2}{11}-\frac{2}{11}\right)\)

\(+\left(\frac{7}{13}-\frac{7}{13}\right)-\frac{9}{16}\)

\(=0+0+0+0-\frac{1}{16}\)

\(=\frac{-1}{16}\)

a) \(2^5+5.13-3.2^3\)

\(=32+5.13-3.8\)

\(=32+65-24\)

\(=97-24\)

\(=73\)

b) \(5^{13}:5^{10}-5^2.2^2\)

\(=5^3-25.4\)

\(=125-100\)

\(=25\)

c) \(4^5:4^3-3^9:3^7+5^0\)

\(=4^2-3^2+1\)

\(=16-9+1\)

\(=7+1\)

\(=8\)

a)\(\frac{2}{3}\times\frac{4}{5}-\frac{3}{5}\times\frac{1}{2}=\frac{8}{15}-\frac{3}{10}\)

                                            \(=\frac{7}{30}\)

b)         \(\frac{12}{5}-\frac{4}{7}-\frac{3}{7}=\frac{64}{35}-\frac{3}{7}\)

                                            \(=\frac{7}{5}\)

cho chj nhé em THKS em nhìu!

CHÚC EM HỌC TỐT!

a)1/2-7/13-1/3+ -6/12+1/2+1 1/3

  =(1/2-1/2)+(1 1/3 - 1/3)-6/12+7/13

  =0+1-1/2+7/13

 =1-1/2+7/13

 =1/2+7/13

=27/26

b)0,75+2/5+1/9 - 1 2/5 + 5/4

   =3/4+2/5+1/9-1 2/5+5/4

    =(3/4+5/4)-(1 2/5-2/5)+1/9

     =2-1+1/9

     =1+1/9

     =10/9

     c) (-1 1/2:3/4).(-4 1/2)-1/2

        =(-1,5:0,75).(-4,5)-0,5

        =-2.-4,5-0,5

        =9-0,5

        =8,5

a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)

\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)

\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

nêu viết đề bài như thế thì nghĩ như là\(\frac{2\cdot4}{1}+\frac{2\cdot7}{4}...\)
=> hai bạn kia sai

A = 0-1 + 2-3 + 4-5 +...+ 2017-2018 

=> A = (-1) + (-1) + (-1)  +...+ (-1) (Có 1009 số hạng)

=> A = 1009.(-1)

=> A = -1009

B = 1-3+5-7+ 9-11+....+2005-2007 

=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)

=> B = 502.(-2)

=> B = -1004

C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102

=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)

=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)

=> C = (-9).17

=> C = -153

cho gia tri tong a,b,c

Bài 1: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) (1)

Từ \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)

Thay vào (1) ta có:

\(\frac{a^2+ab}{b^2+ab}=\frac{a}{b}\Rightarrow\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\) (luôn đúng)

Vậy ta có điều phải chứng minh

Bài 2: 

\(B=\dfrac{2^{15}\cdot5^8-2^5\cdot2^9\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{5}{4}\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)