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câu nào dạng cũng giống nhau, ko biết 1 câu là ko giải đc toàn bộ
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\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{2}+\frac{3}{4}+\frac{3}{4}\right)-2=?\)
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\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{2}-\frac{1}{50}\)
=\(\frac{12}{25}\)
Dấu chấm là dấu nhân,bạn bít rồi đúng ko
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}=\frac{25}{50}-\frac{1}{50}=\frac{24}{50}=\frac{12}{25}\)
Công thức : \(\frac{a}{b\left(b+a\right)}=\frac{1}{b}-\frac{1}{b+a}\)
\(\frac{2a}{b\left(b+a\right)\left(b+2a\right)}=\frac{1}{b\left(b+a\right)}-\frac{1}{\left(b+a\right)\left(b+2a\right)}\)
\(\frac{3a}{b\left(b+a\right)\left(b+2a\right)\left(b+3a\right)}=\frac{1}{b\left(b+a\right)\left(b+2a\right)}-\frac{1}{\left(b+a\right)\left(b+2a\right)\left(b+3a\right)}\)
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B= 1/1.2.3+1/2.3.4+....+1/48.49.50
B= 1/1.2-1/2.3+1/2.3-1/3.4+...+1/48.49-1/49.50
B = 1/1.2-1/49.50
B= 1/2-1/2450
B= 612/1225