Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)
\(=0-0+0-0+\dfrac{10}{9}\)
\(=\dfrac{10}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)
\(=1+1+\dfrac{3}{105}\)
\(=\dfrac{213}{105}=\dfrac{71}{35}\)
Câu 1 dễ mà :
1.2.3...9 - 1.2.3...8 - 1.2.3...7.82
= 1.2.3...8.9 - 1.2.3...8.1 - 1.2.3...7.8.8
= 1.2.3...8.( 9 - 1 - 8 )
= 1.2.3...8.0
= 0
\(\frac{5}{9}-\frac{7}{13}-\frac{5}{9}.\frac{3}{13}+\frac{-5}{9}.\frac{6}{9}\)
= \(\frac{5}{9}-\frac{7}{13}-\frac{5}{3}.\frac{1}{13}+\frac{-5}{3}.\frac{2}{9}\)
=\(\frac{5}{9}-\frac{7}{13}-\frac{5}{39}+\frac{-10}{27}\)
=\(\frac{5}{9}+\frac{-7}{13}+\frac{-5}{39}+\frac{-10}{27}\)
=\(\frac{-13}{27}\)
Chúc bạn học tốt nhea ♥♥♥
\(\frac{5}{9}-\frac{7}{13}-\frac{5}{9}.\frac{3}{13}+\frac{-5}{9}.\frac{61}{9}\)
\(=\frac{5}{9}\left(1-\frac{3}{13}\right)-\frac{7}{13}-\frac{5}{9}.\frac{61}{9}\)
\(=\frac{5}{9}.\frac{10}{13}-\frac{5}{9}.\frac{61}{9}-\frac{7}{13}\)
\(=\frac{5}{9}\left(\frac{10}{13}-\frac{61}{9}\right)-\frac{7}{13}\)\(=-\frac{314}{81}\)
a) $371+731-271-531$
$=(371-271)+(731-531)$
$=100+200$
$=300$
b) $57+58+59+60+61-17-18-19-20-21$
$=(57-17)+(58-18)+(59-19)+(60-20)+(61-21)$
$=40+40+40+40+40$
$=40\cdot5$
$=200$
c) $9-10+11-12+13-14+15-16$
$=(9-10)+(11-12)+(13-14)+(15-16)$
$=-1+(-1)+(-1)+(-1)$
$=-1\cdot4$
$=-4$
$\text{#}Toru$
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
d)
đặt A = 1 + 2 + 22 + ... + 280
2A = 2 + 22 + 23 + ... + 281
2A - A = ( 2 + 22 + 23 + ... + 281 ) - ( 1 + 2 + 22 + ... + 280 )
A = 281 - 1 > 281 - 2
e)
đặt \(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{899}{900}\)
\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{900}\right)\)
\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)
\(A=29-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\right)\)
đặt \(B=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{900}\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{30^2}\)
\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{29}-\frac{1}{30}\)
\(=1-\frac{1}{30}=\frac{29}{30}< 1\)
\(\Rightarrow A< 29\)
So sánh C và D biết
C=1+13+13^2+...+13^13/1+13+13^2+...+13^12
D=1+11+11^2+...+11^13/1+11+11^2+...+11^12
\(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=\frac{1}{9}\)
\(\frac{2^{10}\cdot3^{13}\cdot16^3}{4^{10}\cdot9^6}=\frac{2^{10}\cdot3^{13}\cdot2^{12}}{2^{20}\cdot3^{12}}=\frac{2^{22}\cdot3^{13}}{2^{20}\cdot3^{12}}=2^2\cdot3=12\)
9/(7.10) + 9/(10.13) + 9/(13.16) + ... + 9/(58.61)
= 3.(1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16 + ... + 1/58 - 1/61)
= 3.(1/7 - 1/61)
= 3 . 54/427
= 162/427