1+1×11-1×1÷1-1+1+11+1-13=
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Bài 1:
\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)
\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)
\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)
=1/131
Bài 2:
b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)
\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)

a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79
= 1/10 - 1/79
= máy tính ok
mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha
a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)
b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)
c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)
d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)
e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)
g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)

B=1/4x (7+9) + 1/5x(9+11) + 1/6x(11+13) + 1/7x13
=4+2+4+\(\frac{13}{7}\)=10+\(\frac{13}{7}\)=\(\frac{83}{7}\)


a) \(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
=\(\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)-\left(\frac{3}{4}+\frac{2}{9}+\frac{1}{36}\right)+\frac{1}{64} \)
=\(\frac{5+9+1}{15}-\frac{27+8+1}{36}+\frac{1}{64}\)
= \(1+1+\frac{1}{64}=2\frac{1}{64}\)
1 + 1 × 11 - 1 × 1 ÷ 1 - 1 + 1 + 11 + 1 - 13
= 1 + 11 - 1 - 1 + 1 + 11 + 1 - 13
= 12 - 1 - 1 + 1 + 11 + 1 - 13
= 11 - 1 + 1 + 11 + 1 - 13
= 10 + 1 + 11 + 1 - 13
= 11 + 11 + 1 - 13
= 22 + 1 - 13
= 23 - 13
= 10