\(\left(\frac{x-10}{1994}-1\right)\)+\(\left(\frac{x-8}{1996}-1\right)\)+\(\left(\frac{x-6}{1998}-1\right)\)+\(\left(\frac{x-4}{2000}-1\right)\)+\(\left(\frac{x-2}{2002}-1\right)\)=\(\left(\frac{x-2002}{2}-1\right)\)+\(\left(\frac{x-2000}{4}-1\right)\)+\(\left(\frac{x-1998}{6}-1\right)\)+\(\left(\frac{x-1996}{8}-1\right)\)+\(\left(\frac{x-1994}{10}-1\right)\)

suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)=\(\frac{x-2004}{2}\)+\(\frac{x-2004}{4}\)+\(\frac{x-2004}{6}\)+\(\frac{x-2004}{8}\)+\(\frac{x-2004}{10}\)

suy ra  \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)- \(\frac{x-2004}{2}\)- \(\frac{x-2004}{4}\)- \(\frac{x-2004}{6}\)- \(\frac{x-2004}{8}\)- \(\frac{x-2004}{10}\)=0

suy ra (x-2004) . ( \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\))=0

Vì  \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\) khác 0

nên x-2004=0 suy ra x=2004

em cảm ơn

\(C=\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\right)-\)\(\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}\right)\)

\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+..+\frac{1}{2017}\)

\(\Rightarrow C=\left(\frac{1}{101}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\right)-\left(\frac{1}{1010}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)

\(\Rightarrow C=\frac{1}{2018}\)

B=\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\).....\(\left(1-\frac{1}{1+2+3+4+...+100}\right)\)

B= \(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).....\)\(\left(1-\frac{1}{100.101:2}\right)\)

B= \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.....\frac{5049}{5050}\)

B= \(\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}.....\frac{5049.2}{5050.2}\)

B= \(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{99.102}{100.101}\)

B=\(\frac{1.2.3.4.....99}{2.3.4.5.....100}.\frac{4.5.6.7.....102}{3.4.5.6.....101}\)

B=\(\frac{1}{100}.\frac{102}{3}=\frac{17}{50}\)

Ta có 3a-b+ab=8
=>a.(3+b)-(3+b)=5(trừ hai về ik 3)
=>(a-1).(3+b)=5
Do a,b là số nguyên dương nên a-1 và b+3 là cặp ước của 5

tự lập bảng làm nhé

Ta có : 3a -b + ab = 8

=> a.( 3 + b ) - ( 3 + b ) = 5 

=> ( a - 1 ) . ( 3 + b ) = 5

Do a và b là số nguyên dương nên a-1 và b+3 là cặp ước chung của 5

Ti vi loại 49 inh có đường chép màn hình là :

 2,54 . 49 = 124 ( cm )

Vậy : . . 

HOK TỐT

từ đồng nghĩa: tháng giêng-tháng đầu ;ai-người ta 

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