Question

(Theo dạng dãy số có quy luật)

a) Tính A= \(\frac{1}{2}-\frac{1}{3}\); B= \(\frac{1}{3}-\frac{1}{4}\); C= \(\frac{1}{4}-\frac{1}{5}\)

b) Tính A + B và A+B+C

Answer 1

\(a,Tính\)

\(A=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(B=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\)

\(C=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\)

\(b,Tính\)

\(A+B=\left[{}\begin{matrix}\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\\\frac{1}{6}+\frac{1}{12}=\frac{1}{4}\end{matrix}\right.\) ( 2 cách )

\(A+B+C=\left[{}\begin{matrix}\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}\\\frac{1}{6}+\frac{1}{12}+\frac{1}{20}=\frac{3}{10}\end{matrix}\right.\) ( 2 cách )

Answer 2

a)

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

\(A\) \(=\) \(\frac{1}{6}\).

\(B=\frac{1}{3}-\frac{1}{4}\)

\(B=\frac{1}{12}\).

\(C=\frac{1}{4}-\frac{1}{5}\)

\(C=\frac{1}{20}\).

b)

\(A+B=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\)

\(A+B=\frac{1}{6}+\frac{1}{12}\)

\(A+B=\frac{1}{4}\).

\(A+B+C=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)\)

\(A+B+C=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(A+B+C=\frac{3}{10}\).

Chúc bạn học tốt!