Mk nghĩ là thê này nè , k biêt co đung k:

Ta co \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\) (1)

\(.......\)

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

Từ (1) \(=>B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}\)\(=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)

\(=>B< 1\left(đpcm\right)\)

tick cho mk nha

a/ => x³ = 64 => x³ = 4³ => x = 4

b/ => x⁵ = 32 => x⁵ = 2⁵ => x = 2 

c/ => 3^x = 81 => 3^x = 3⁴ => x = 4

d/ => 3^x+2 = 243 => 3^x+2 = 3⁵ => x + 2 = 5 => x = 3

e/ => (x - 2)³ = 64 => (x - 2)³ = 4³ => x - 2 = 4 => x = 6

f/ => (x - 8)² = 81 => (x - 8)² = 9² => x - 8 = 9 => x = 17

Ta co : \(\dfrac{2k+2}{k-3}=\dfrac{2k-6+8}{k-3}=\dfrac{2.\left(k-3\right)+8}{k-3}=\dfrac{2.\left(k-3\right)}{k-3}+\dfrac{8}{k-3}=2+\dfrac{8}{k-3}\)

Để \(\dfrac{2k+2}{k-3}\) nguyên thì \(\dfrac{8}{k-3}\) cũng phải nguyên

\(=>k-3\inƯ\left(8\right)\)

\(=>k-3\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(=>k\in\left\{-5;-1;1;2;4;5;7;11\right\}\)

tick cho mk nha

chưa hiểu chỗ nào thì hỏi mk nha

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)

\(=>\dfrac{1}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(=>\dfrac{1}{3}.\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(=>\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.3\)

\(=>\dfrac{x+3-5}{5.\left(x+3\right)}=\dfrac{303}{1540}\)

\(=>\dfrac{x-2}{5x+15}=\dfrac{303}{1540}\)

\(=>1540.\left(x-2\right)=303.\left(5x+15\right)\)

\(=>1540x-3080=1515x+4545\)

\(=>1540x-1515x=3080+4545\)

\(=>25x=7625\)

\(=>x=305\)

Vậy x = 305

tick cho mk nha

co gì chưa hiểu thì hỏi nha

Co 3 trường hợp:

TH1: a > b

Ta co \(\dfrac{a}{b}-1=\dfrac{a-b}{b}\)

\(\dfrac{a+n}{b+n}-1=\dfrac{a+n-b+n}{b+n}=\dfrac{a-b}{b+n}\)

Vì \(\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\) nên \(\dfrac{a}{b}-1>\dfrac{a+n}{b+n}-1\)

Hay \(\dfrac{a}{b}>\dfrac{a+n}{b+n}\) khi a > b

TH2: a < b

Ta co \(1-\dfrac{a}{b}=\dfrac{b-a}{b}\)

\(1-\dfrac{a+n}{b+n}=\dfrac{b+n-a+n}{b+n}=\dfrac{b-a}{b+n}\)

Vì \(\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\)nên \(1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\)

Hay \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\)khi a < b

TH3 : a = b

\(\dfrac{a+n}{b+n}=\dfrac{a}{b}=1\)

Vậy \(\dfrac{a}{b}=\dfrac{a+n}{b+n}\) khi a = b

tick cho mk nha

co gì chưa đung thì xem lại rồi hỏi mk

\(A=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}\)

\(A=\left(\dfrac{-9-2-7}{18}\right)+\left(\dfrac{21+4+10}{35}\right)+\dfrac{1}{127}\)

\(A=-1+1+\dfrac{1}{127}\)

\(A=\dfrac{1}{127}\)

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.4}+\dfrac{1}{3.4.5.4}+...+\dfrac{1}{98.99.100.4}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.\left(5-1\right)}+\dfrac{1}{3.4.5.\left(6-2\right)}+...+\dfrac{1}{98.99.100.\left(101-97\right)}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5-1.2.3.4}+\dfrac{1}{3.4.5.6-2.3.4.5}+...+\dfrac{1}{98.99.100.101-97.98.99.100}\)

\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}-\dfrac{1}{1.2.3.4}+\dfrac{1}{3.4.5.6}-\dfrac{1}{2.3.4.5}+...+\dfrac{1}{98.99.100.101}-\dfrac{1}{97.98.99.100}\)

\(\dfrac{1}{4}B=\dfrac{1}{98.99.100.101}\)

\(B=\dfrac{1}{98.99.100.101}.4=\dfrac{1}{98.99.25.101}\)

tick cho mk nha

bài tự làm 100%

co gì chưa đc thì coi lại nha

1/Ta co :

\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}\right)+\dfrac{46}{45}.\left(\dfrac{15}{7}-\dfrac{45}{46}\right)\)

=\(\dfrac{15}{7}.\dfrac{1}{5}-\dfrac{15}{7}.\dfrac{46}{45}+\dfrac{46}{45}.\dfrac{15}{7}-\dfrac{46}{45}.\dfrac{45}{46}\)

=\(\dfrac{15}{7}.\left(\dfrac{1}{5}-\dfrac{46}{45}+\dfrac{46}{45}\right)-1\)

=\(\dfrac{15}{7}.\dfrac{1}{5}-1\)

=\(\dfrac{3}{7}-1=\dfrac{-4}{7}\)

2/Ta co

\(\dfrac{43}{47}.\left(\dfrac{18}{37}+\dfrac{47}{43}\right)-\dfrac{18}{3}.\left(\dfrac{43}{47}+\dfrac{37}{36}\right)\)

=\(\dfrac{43}{47}.\dfrac{18}{37}+\dfrac{43}{47}.\dfrac{47}{43}-\dfrac{18}{37}.\dfrac{43}{47}+\dfrac{18}{37}.\dfrac{37}{36}\)

=\(\dfrac{18}{37}.\left(\dfrac{43}{37}-\dfrac{43}{37}+\dfrac{37}{36}\right)+1\)

=\(\dfrac{18}{37}.\dfrac{37}{36}+1\)

=\(\dfrac{1}{2}+1=\dfrac{3}{2}\)

tick cho mk nha