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a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\)
=> \(A=\frac{9}{10}\)
b/ \(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}\)
=> \(A=1+\frac{7}{n-5}\)
Để A nguyên => 7 chia hết cho n-5 => n-5=(-7; -1; 1; 7)
=> n=(-2; 4, 6, 8)
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a=1/1.2+1/2.3+1/3.4+...............+1/7.8 ta thấy hiệu cuẩ mẫu bằng tử suy ra
a=1-1/2+1/2-1/3+...................+1/6-1/7+1/7-1/8
sau khi giảm ước đi ta còn 1-1/8=7/8
kết quả là 7/8
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1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
=1-1/9
=8/9
A) \(x = {7 \over 10}- {8 \over10} \)
\(x = {-1 \over 10}\)
B)\({2 \over3}x = 2{5 \over 6}-{3 \over4}\)
\({2 \over3}x = {25 \over 12}\)
\(x = {25 \over 12}/{2 \over3} \)
\(x = {25\over 8}\)
2/ Tính tổng:
\( = {8 \over 9}\)
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\(a)\frac{62}{7}\cdot x=\frac{29}{9}\div\frac{3}{56}\)
\(\Rightarrow\frac{62}{7}\cdot x=\frac{29}{9}\cdot\frac{56}{3}\)
\(\Rightarrow\frac{62}{7}\cdot x=\frac{1624}{27}\)
\(\Rightarrow x=\frac{1624}{27}\div\frac{62}{7}\)
\(\Rightarrow x=\frac{1624}{27}\cdot\frac{7}{62}\)
\(\Rightarrow x=\frac{11368}{1674}=\frac{5684}{837}\)
Rút gọn thử đi
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a, b phân k ra là ok
\(c)\) \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(A=\left(-1\right)\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=\left(-1\right).\frac{3}{20}\)
\(A=\frac{-3}{20}\)
Vậy \(A=\frac{-3}{20}\)
Chúc bạn học tốt ~
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\(A=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}+...+\frac{1}{n^2}-\frac{1}{4n}=\frac{56}{673}\)
\(\Rightarrow4A=\)
\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n\left(n+4\right)}=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)=\frac{56}{673}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{56}{673}:\frac{1}{4}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{3}-\frac{224}{673}\)
\(\Rightarrow\frac{1}{n+4}=\frac{1}{2019}\)
=> n + 4 = 2019
n = 2019 - 4
n = 2015