BN
1
K
Khách
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TP
2
24 tháng 4 2020
(1 - 1/2^2)(1 - 1/3^2).......(1 - 1/10^2)
= ( 2^2 -1/2^2)(3^2-1/3^2).......(10^2-1/10^2)
= (4-1/2.2)(9-1/3.3)......(100-1/10.10)
=3/2.2 . 8/3.3..........99/10.10
=1.3/2.2 . 2.4/3.3........9.11/10.10
=(1/2 . 2/3 . ........ 9/10)(3/2 . 4/3......... 11/10)
= 1/10 . 11/2
= 11/20
24 tháng 4 2020
A=(1-1/22)(1-1/32).......(1-1/102)
=(1-1/4)(1-1/9)......(1-1/100)
=3/4.8/9......99/100
=1.3.2.4...9.11/(2.3....10)2
=1.2.3....11/(3...10).10
=2.11/10
=22/10
=11/5
NX
16 tháng 8 2023
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
NX
16 tháng 8 2023
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
NM
29 tháng 7 2023
62 . 10 : { 780 : [ 103 - ( 2 . 53 + 35 . 14 ) ] }
= 360 : { 780 : [ 1000 - ( 250 + 490 ) ] }
= 360 : { 780 : [ 1000 - 740 ] }
= 360 : { 780 : 260 }
= 360 : 3
= 120.
31 tháng 12 2017
Bài 2:
a)|x| < 3
x\(\in\){-2;-1;0;1;2}
b)|x - 4 | < 3
x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }
c) | x + 10 | < 2
x\(\in\){ -2 ; -10 }
EG
31 tháng 12 2017
Bài 1:
A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99
A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]
A = 1617 + [(-1)+(-1)+...+(-1)]
A = 1617 + (-49)
A = +(1617-49) = A = 1568
B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60
B =
2)
a) \(x\in\left\{2;1;0;-1;-2\right\}\)
b) \(x\in\left\{6;-6;5;-5;4\right\}\)
c) \(x\in\left\{-9;-11;-10\right\}\)
3)
\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)
VN
Võ Ngọc Phương
VIP
23 tháng 8 2023
a) \(\dfrac{-15}{4}:\dfrac{21}{-10}=\dfrac{-15}{4}.\dfrac{-10}{21}=\dfrac{25}{14}\)
b) \(\dfrac{-7}{14}:\left(-0,14\right)=\dfrac{-7}{14}.\dfrac{-50}{7}=\dfrac{25}{7}\)
c) \(\left(\dfrac{-11}{15}\right):1\dfrac{1}{10}=\dfrac{-11}{15}.\dfrac{10}{11}=\dfrac{-2}{3}\)
d) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)
BD
23 tháng 8 2023
\(a.-\dfrac{15}{4}:\left(\dfrac{21}{-10}\right)\)
\(=-\dfrac{15}{4}\cdot\left(-\dfrac{10}{21}\right)\)
\(=\dfrac{25}{14}\)
\(b.-\dfrac{7}{14}:\left(-0,14\right)\)
\(=-\dfrac{1}{2}:\left(-\dfrac{7}{50}\right)\)
\(=\dfrac{25}{7}\)
\(c.\left(-\dfrac{11}{15}\right):\left(1\dfrac{1}{10}\right)\)
\(=\left(-\dfrac{11}{15}\right):\dfrac{11}{10}\)
\(=-\dfrac{2}{3}\)
\(d.\left(2\dfrac{1}{7}\right):\left(1\dfrac{1}{14}\right)\)
\(=\dfrac{15}{7}:\dfrac{15}{14}\)
\(=2\)
NX
16 tháng 8 2023
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
NX
16 tháng 8 2023
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
23 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{19}\)
`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{19}{19}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
`b)`
\(-\dfrac{3}{5}\cdot\dfrac{5}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{6}{7}\)
`=`\(-\dfrac{3}{5}\cdot\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
`=`\(-\dfrac{3}{5}\cdot\dfrac{14}{7}\)
`=`\(-\dfrac{3}{5}\cdot2=-\dfrac{6}{5}\)
`c)`
\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
`=`\(10\dfrac{2}{9}+2\dfrac{2}{5}-7\dfrac{2}{9}\)
`=`\(\left(10\dfrac{2}{9}-7\dfrac{2}{9}\right)+2\dfrac{2}{5}\)
`=`\(3+2\dfrac{2}{5}=\dfrac{27}{5}\)
`d)`
\(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
`=`\(6\dfrac{3}{10}-3\dfrac{4}{7}-2\dfrac{3}{10}\)
`=`\(\left(6\dfrac{3}{10}-2\dfrac{3}{10}\right)-3\dfrac{4}{7}\)
`=`\(4-3\dfrac{4}{7}=\dfrac{3}{7}\)
VN
Võ Ngọc Phương
VIP
23 tháng 7 2023
a) \(\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)
\(=\dfrac{7}{5}.\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)
\(=\dfrac{7}{5}.1\)
\(=\dfrac{7}{5}\)
b) \(\dfrac{-3}{5}.\dfrac{5}{7}+\dfrac{-3}{5}.\dfrac{3}{7}+\dfrac{-3}{5}.\dfrac{6}{7}\)
\(=\dfrac{-3}{5}.\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)
\(=\dfrac{-3}{5}.2\)
\(=\dfrac{-6}{5}\)
c) \(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)
\(=\dfrac{92}{9}+\dfrac{12}{5}-\dfrac{65}{9}\)
\(=\dfrac{92}{9}-\dfrac{65}{9}+\dfrac{12}{5}\)
\(=3+\dfrac{12}{5}\)
\(=\dfrac{15}{5}+\dfrac{12}{5}\)
\(=\dfrac{27}{5}\)
d) \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)
\(=\dfrac{63}{10}-\dfrac{25}{7}-\dfrac{23}{10}\)
\(=\dfrac{63}{10}-\dfrac{23}{10}-\dfrac{25}{7}\)
\(=4-\dfrac{25}{7}\)
\(=\dfrac{28}{7}-\dfrac{25}{7}\)
\(=\dfrac{3}{7}\)
Chúc bạn học tốt
KT
2
AH
Akai Haruma
Giáo viên
9 tháng 10 2023
Lời giải:
Ta có:
$10\equiv -1\pmod {11}$
$\Rightarrow 10^{2022}\equiv (-1)^{2022}\equiv 1\pmod {11}$
$\Rightarrow A=10^{2022}-1\equiv 1-1\equiv 0\pmod {11}$
Vậy $A\vdots 11$
TL
23 tháng 12 2023
ok
A= 10^2022-1
Ta có thể thấy 10^2022=100000000...........0000000000
10000000.......0000000000-1 thì lúc nnày tổng bằng
9999999999999999........................999999999999999999999
mà 99999999999999999999999....................9999999999999999999chia hết cho 11 nên tổng này chia hết cho 11
BA
5 tháng 10 2019
\(7^{13}:49^2=7^{13}:7^4=7^9\)
\(27^{16}:9^{10}=3^{48}:3^{20}=3^{28}\)
\(5^{20}\cdot9^{10}=5^{20}\cdot3^{20}=15^{20}\)
\(7^7\cdot13+7^7\cdot36=7^7\cdot\left(13+36\right)=7^7\cdot49=7^7\cdot7^2=7^9\)
\(5^{12}\cdot37-5^{12}\cdot12=5^{12}\cdot\left(37-12\right)=5^{12}\cdot25=5^{12}\cdot5^2=5^{14}\)
\(A=10^2\times\left[\left(-2\right).10,1-20,21+2.10,11\right]\)
\(=100\times\left[\left(-20,2\right)-20,21+20,22\right]\)
\(=-20,19\times100=-2019\)