A=9217/20234    B=âm 5930,096468

Bạn ơi cách giải bạn ak

\(\frac{\text{1984.1985 + 1986.20 + 1965}}{1985.200-1984.1986}=-1,123249617\approx-1,13\)

hãy rút gọn phân số

Kkết quả là:

-1.123249617

-\(\frac{1989.1985+1986.20+1965}{1985.2000-1984.1985}\)

\(=\frac{1989.1985+1986.20-20+1985}{1985\left(2000-1984\right)}\)

\(=\frac{1985\left(1989+20+1\right)}{1985.16}\)

\(=\frac{2010}{16}=\frac{1005}{8}=125,625\)

Anh chị cứu em

:V toán lp 3 cơ ak 

A = \(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{4347}\)

\(A\cdot3=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1449}\)

\(A\cdot3-A=\left(\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1449}\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{18}+...+\frac{1}{4347}\right)\)

\(A\cdot2=\frac{3}{2}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{1449}-\frac{1}{2}-\frac{1}{6}-\frac{1}{18}-...-\frac{1}{4347}\)

\(A\cdot2=\frac{3}{2}-\frac{1}{4347}\)

\(A\cdot2=\frac{13039}{8694}\)

\(A=\frac{13039}{8694}:2\)

\(A=\frac{13039}{17388}\)

Kết quả hơi lớn nên kiểm tra lại đề :))

\(1984\cdot1985+1986\cdot20+\frac{1965}{1985\cdot2000}-1984\cdot1985\)

\(=3938240+39720+\frac{393}{794000}-3938240\)

\(=3938240-3938240+39720+\frac{393}{794000}\)

\(=39720+\frac{393}{794000}\)

\(=39720,00049\)rồi bạn tự đổi ra phân số nhé

ý câu hỏi là thế này mà bạn :

\(\frac{\text{1984.1985+1986.20+1965}}{\text{1985.2000-1984.1985}}\)

Bài 1:

A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

= \(1-\frac{1}{50}=\frac{49}{50}\)

Bài 2:

Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)

Vậy A < 2

Bài 3:

\(A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)

\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)

Bài 4:

\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)

\(2S=6+3+\frac{3}{2}+...+\frac{3}{2^8}\)

\(2S-S=\left(6+3+\frac{3}{2}+...+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\right)\)

\(S=6-\frac{3}{2^9}=6-\frac{3}{512}=\frac{3069}{512}\)

A=1-1/2+1/2-1/3+.............................1/49-1/50

A=1-1/50

A=49/50

Bài 1 và bài 2 bạn tự làm nhé! Hoặc là search lên mạng tìm vì mik nhớ đã làm nhiều lần lắm rồi! Nhưng nếu bạn cần giải lại thì nhớ nói với mik nha

3)\(A=\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(A=\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
4)\(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^9}\)
\(\Rightarrow\frac{1}{2}S=\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{10}}\)
\(\Rightarrow S=\left(s-\frac{1}{2}S\right).2=\left(3-\frac{3}{2^{10}}\right).2=\frac{3069}{512}\)

A=\(\frac{2001.2002+4002}{2001.2004}\)

A=\(\frac{2001.2002+2001.2}{2001.2004}\)

A=\(\frac{2001.\left(2002+2\right)}{2001.2004}\)

A=\(\frac{2001.2004}{2001.2004}\)=1

k mik nhé

k mk đi

ai k mk

mk k lại

thanks