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2+3/4=11/4 ; 2-3/5=7/5
5/7:6=5/42 ; 5+2/7=37/7
Bài 2: a) x-1/6=1/2+1/3
x-1/6=5/6
x=5/6+1/6
x=1
b) x+1/8=1/2-1/3
x+1/8=1/6
x=1/6-1/8
x=1/24
c) x:1/2=2/3.3/4
x:1/2=1/2
x=1/2.1/2
x =1/4
(x+1)+(x+2)+...x+28=155
=> x + 1 + x + 2 + ... + x + 28 =155
=> 28x + (1 + 2 + 3 + ... + 28) =155
=> 28x + 28.(28+1):2 = 155
=> 28x + 406 = 155
=> 28x = 155 - 406 = -251
=> x = -251/28
Ta có:
(x+1)+(x+2)+...x+28=155
<=> x + 1 + x + 2 + ... + x + 28 =155
<=> 28x + (1 + 2 + 3 + ... + 28) =155
<=> 28x + 28.(28+1):2 = 155
<=> 28x + 406 = 155
<=> 28x = 155 - 406 = -251
=> x = \(-\frac{251}{28}\)
hok tốt
( 1 + 2 + 4 + 8+ ... + 512 ) x ( 101 x 102 - 101 x 101 - 50 - 51 )
= ( 1 + 2 + 4 + 8+ ... + 512 ) x ( 101 x 102 - 101 x 101 - 101 )
= ( 1 + 2 + 4 + 8+ ... + 512 ) x ( 101 - 101 )
= ( 1 + 2 + 4 + 8+ ... + 512 ) x 0
= 0
( 1 + 2 + 4 + 8+ ... + 512 ) x ( 101 x 102 - 101 x 101 - 50 - 51 )
= ( 1 + 2 + 4 + 8+ ... + 512 ) x ( 101 x 102 - 101 x 101 - 101 )
= ( 1 + 2 + 4 + 8+ ... + 512 ) x ( 101 - 101 )
= ( 1 + 2 + 4 + 8+ ... + 512 ) x 0
= 0
\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)
\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)
....
Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được
Sửa đề\(\frac{1}{2x-2021}-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{220}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{1}{2}.\frac{10}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}-\frac{5}{11}=\frac{6}{11}\)
=> \(\frac{1}{2x-2021}=1\)
=> 2x - 2021 = 1
=> 2x = 2022
=> x = 1011
Vậy x = 1011
Ta có : \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
=> \(\frac{1.2.3.....9}{2.3.4....10}=\frac{x}{2010}\)
=> \(\frac{1}{10}=\frac{x}{2010}\)
=> x = 2010/10
=> x = 201
a) Ta có 1+2+3+4+...+x=210
=> x.(x-1):2=210
=>x.(x-1)=210.2=420=20.21
=>x=20
b) Ta có (x+1)+(x+2)+(x+3)+...+(x+99)=6138
=> x.99 +1+2+3+...+99=6138
=>x.99+4851=6138
=>x.99=6138-4851=1287
=>x=1287:99=13
(1+1/2)(1+1/3)...(1+1/99)
=(3/2)(4/3)(5/4)...(100/99)
=100/2=50
Bằng 1/99 mà bạn
50 k cho mình nha
Thật ko ớ. Nếu thật mình k cho. Mà nếu đúng thì cậu giải thích rõ cho mình nha
bn tự chép đầu bài nhé
( 2/2 + 1/2 ) x ( 3/3 + 1/3 ) +( 4/4 + 1/4 ) + ... + ( 99/99 + 1/99 )
= 3/2 x 4/3 x 5/4 x ... x 100/99
= 3 x 4 x 5 x ... x 100/ 2 x 3 x 4 x ... x 99
= ( 3 x 4 x 5 x ... x 99 ) x 100 / 2 x ( 3 x 4 x ... x 99 )
= 100 / 2
= 50
\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)......\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{100}{99}\)
\(=\frac{1}{2}.100\)
\(=50\)