K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

26 tháng 1 2016

SSH của B : ( 3997 - 1 ) : 2 +1 = 1999

=> B = ( 3997 + 1 .1999 : 2 = 3996001

Vậy B = 3996001

26 tháng 1 2016

Lấy số cuối trừ số đầu chia khoảng cách cộng 1 . áp dụng công thức ấy mà tick . Tick mình nha !

10 tháng 6 2017

a) = 29/15

b) = 7/15

c) = 1

d) = 3

e) = 67/17

f) = 2

mk nhanh nhất tk cho mk nha

10 tháng 6 2017

a/\(\frac{3}{5}+\frac{4}{3}=\frac{9}{15}+\frac{20}{15}=\frac{29}{15}\)

b/\(\frac{2}{3}-\frac{1}{5}=\frac{10}{15}-\frac{3}{15}=\frac{7}{15}\)

c/\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{7}{6}\)

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

1)

\(\left(a\right)37+397+3997+39997\)

\(=40-3+400-3+4000-3+40000-3\)

\(=\left(40+400+4000+40000\right)-\left(3+3+3+3\right)\)

\(=44440-12=44428\)

\(\left(b\right)298+2998+29998+299998\)

\(=300-2+3000-2+30000-2+300000-2\)

\(=\left(300+3000+30000+300000\right)-\left(2+2+2+2\right)\)

\(=333300-8=333296\)

\(\left(c\right)9+99+999+9999+99999\)

\(=10-1+100-1+1000-1+10000-1+100000-1\)

\(=\left(10+100+1000+10000+100000\right)-\left(1+1+1+1+1\right)\)

\(=111110-5=111105\)

2)
\(\left(a\right)\left(2+4+6+...+2002+2004+2006\right)-\left(1+3+5+...+2001+2003+2005\right)\)

\(=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(2002-2001\right)+\left(2004-2003\right)+\left(2006-2005\right)\)

\(=1+1+1+...+1+1+1\)( 1003 số 1 )

\(=1003\)

\(\left(b\right)88-87+86-85+84-83+...+6-5+4-3+2-1\)

\(=\left(88-87\right)+\left(86-85\right)+\left(84-83\right)+...+\left(6-5\right)+\left(4-3\right)+\left(2-1\right)\)

\(=1+1+1+...+1+1+1\)( 44 số 1 )

\(=44\)

\(\left(c\right)100-98+96-94+92-90+...+12-10+8-6+4-2\)

\(=\left(100-98\right)+\left(96-94\right)+\left(92-90\right)+...+\left(12-10\right)+\left(8-6\right)+\left(4-2\right)\)

\(=2+2+2+...+2+2+2\) ( 25 số 2 )

\(=50\)

3)

\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)

\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)

\(=3+3+3+3+3+3+3+3+3+3+3=33\)

\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)

\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)

\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)

\(=2006-1275=731\)

\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)

\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)

\(=4+4+4+4+4+4+4+4+4+4+4=44\)

23 tháng 7 2020

Cảm ơn bạn nha

28 tháng 1 2022

a)\(\frac{-3}{5}+\frac{1}{4}+\frac{-3}{10}\)

\(=\frac{-12}{20}+\frac{5}{20}+\frac{-6}{20}\)

\(=\frac{-13}{20}\)

b)\(\frac{1}{5}+\frac{-9}{10}+\frac{-7}{25}=\frac{10}{50}+\frac{-45}{50}+\frac{-14}{50}=\frac{-49}{50}\)

c)\(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)

=\(\left(\frac{1}{5}+\frac{-1}{5}\right)+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{1}{7}+\frac{-1}{7}\right)+\left(\frac{1}{8}+\frac{-1}{8}\right)+\frac{1}{9}\)

=\(0+0+0+0+\frac{1}{9}\)

=\(\frac{1}{9}\)

a, \(\frac{4}{7}x\frac{5}{7}+\frac{3}{7}x\frac{5}{6}\)

= ( \(\frac{4}{7}+\frac{3}{7}\)) x \(\frac{5}{7}x\frac{5}{6}\)

= 1 x \(\frac{5}{7}x\frac{5}{6}\)

\(\frac{5}{7}x\frac{5}{6}\)

\(\frac{25}{42}\)

Tương tự mấy câu sau cũng làm như thế này

a)\(2-3+5-7+9-11+13-15+17=\left(2+5+9+13+17\right)-\left(3+7+11+15\right)\)

                                                                                               \(=46-36=10\)

b)\(\frac{1}{1.2}+\frac{1}{2.3}+...............+\frac{1}{8.9}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.................+\frac{1}{8}-\frac{1}{9}\)

                                                                        \(=\frac{1}{1}-\frac{1}{9}=\frac{9}{9}-\frac{1}{9}=\frac{8}{9}\)

Áp dụng \(\frac{1}{n.\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Chúc bạn học tốt

19 tháng 10 2020

Đề là tính bằng cách hợp lý đúng ko bạn

a, 2-3+5-7+9-11+13-15+17

= (5+13) - (3+15) + (2+9-11) + (17-7)

= 18 - 18 + 0 +10

= 10

b, \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}\)

\(=\frac{8}{9}\)

12 tháng 10 2023

a) 3 2/7 - 2/5 + 5/7 - 3/5

= 3 + 2/7 - (2/5 + 3/5) + 5/7

= 3 + (2/7 + 5/7) - 1

= 3 + 1 - 1

= 3

b) 5 4/7 - 4/9 + 1 3/7 - 5/9

= 5 + 4/7 - 4/9 + 1 + 3/7 - 5/9

= (5 + 1) + (4/7 + 3/7) - (4/9 + 5/9)

= 6 + 1 - 1

= 6

12 tháng 10 2023

a) \(3\dfrac{2}{7}-\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{3}{5}\)

\(=\left(\dfrac{23}{7}+\dfrac{5}{7}\right)-\left(\dfrac{2}{5}+\dfrac{3}{5}\right)\)

\(=4-1\)

\(=3\)

b) \(5\dfrac{4}{7}-\dfrac{4}{9}+1\dfrac{3}{7}-\dfrac{5}{9}\)

\(=\left(\dfrac{32}{7}+\dfrac{10}{7}\right)-\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=6-1\)

\(=5\)

14 tháng 6 2015

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

10 tháng 8 2022

 

1.3.77−1​+3.7.99−3​+7.9.1313−7​+9.13.1515−9​+\frac{19-13}{13.15.19}+13.15.1919−13​

=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31​−3.71​+3.71​−7.91​+7.91​−9.131​+9.131​−13.151​+13.151​−15.191​

=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31​−15.191​=28595​−2851​=28594​

b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61​.(1.3.76​+3.7.96​+7.9.136​+9.13.156​+13.15.196​)

làm giống như trên

c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81​.(1.2.31​+2.3.41​+3.4.51​+...+48.49.501​)

=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161​.(1.2.32​+2.3.42​+3.4.52​+...+48.49.502​)

=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161​.(1.2.33−1​+2.3.44−2​+3.4.55−3​+...+48.49.5050−48​)

=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161​.(1.21​−2.31​+2.31​−3.41​+3.41​−4.51​+...+48.491​−49.501​)

=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161​.(21​−24501​)=161​.(24501225​−24501​)=4900153​

d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75​.(1.5.87​+5.8.127​+8.12.157​+...+33.36.407​)

=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75​.(1.5.88−1​+5.8.1212−5​+8.12.1515−8​+...+33.36.4040−33​)

=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75​.(1.51​−5.81​+5.81​−8.121​+8.121​−12.151​+...+33.361​−36.401​)

=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75​.(51​−14401​)=75​.(1440288​−14401​)=28841​

P/S: . là nhân nha