B=1/2.1.2-1/2.2.3+1/2.2.3-1/2.3.4+...+1/2n(n+1)-1/2(n+1)(n+2)

B=1/2[(1/1.2+1/2.3+...+1/n(n+1))-(1/2.3+1/3.4+...+1/(n+1)(n+2))]

Tới đây bạn tự làm tiếp nha, tương tự như bài 1/1.2+1/2.3+..+1/n(n+1) á bạn.Cái này bạn ghi ra bạn sẽ hiểu, mình viết hơi bị lủng củng.

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

Em chỉ làm những bài e biết thôi, thông cảm nhs :D

a/ chịu

b/ \(C=1+7+7^2+.........+7^{50}\)

\(\Leftrightarrow7C=7+7^2+...........+7^{50}+7^{51}\)

\(\Leftrightarrow7C-C=\left(7+7^2+.......+7^{51}\right)-\left(1+7+.....+7^{50}\right)\)

\(\Leftrightarrow6C=7^{51}-1\)

\(\Leftrightarrow C=\dfrac{7^{51}-1}{6}\)

c/ \(A=\dfrac{-1}{4}+\dfrac{7}{3}+\dfrac{3}{4}+\dfrac{9}{2}\)

\(=\left(\dfrac{-1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{7}{3}+\dfrac{9}{2}\right)\)

\(=\dfrac{1}{4}+\dfrac{41}{6}\)

\(=\dfrac{85}{12}\)

d/ Thấy phép tính hơi dài

e/ \(C=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.........+\dfrac{1}{2015.2016.2017}\)

\(\Leftrightarrow2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.........+\dfrac{2}{2015.2016.2017}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.......+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\)

\(=\dfrac{1}{2}-\dfrac{1}{4066272}\)

\(=\dfrac{2033136}{4066272}\)

\(\Leftrightarrow C=\dfrac{2033136}{4066272}:2\)

\(\Leftrightarrow C=?\)

để ý 1+1/x(x+2)=(x²+2x+1)/x(x+2)=(x+1)²/x(x+2)

+ 1+1/1.3=2²/1.3 ;...... 

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+....+\frac{1}{47.48.49.50}\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{47.48.49}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1.2.3}-\frac{1}{48.49.50}\right)\)

\(=\frac{1}{3}.\frac{6533}{39200}=\frac{6533}{117600}\)