A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
= 1-\(\frac{1}{50}\)
= \(\frac{49}{50}\)
ta có công thức tính tổng quát 1/[n(n+1)] = 1/n -1/(n+1)
=> A=1/1.2+ 1/2.3+1/3.4+1/4.5+...+1/49.50
=1/1 -1/2 +1/2 -1/3 +1/3-1/4+.......+1/49 -1/50
= 1 -1/50 = 49/50
Ai thấy đúng thì tk cho mk nhé
= \(\frac{49}{50}\).
Đúng 100% luôn!
Chúc các bạn học giỏi.
A = 1/ 1.2 + 1/2.3 + 1/3.4 + ... + 1/49 .50
= 1 . 1/2 + 1/2 . 1/3 + ... + 1/49 . 1/50
= 1 + 1/50
= 51/50
Ta thấy 1-(1/2*3)=1/1*2+1/2*3
VậyA=1-1,49*50
A=2449/2450
\(\left(\frac{1}{1\cdot2}\right)+\left(\frac{1}{2\cdot3}\right)+\left(\frac{1}{3\cdot4}\right)+...+\left(\frac{1}{49\cdot50}\right)\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{49}-\frac{1}{50}\right)\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+....+\(\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}\)\(-\frac{1}{50}\)=\(\frac{50}{50}-\frac{1}{50}\)
=\(\frac{49}{50}\)
A = \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ .....+ \(\frac{1}{49.50}\)
= 1- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+.......+ \(\frac{1}{49}\)-\(\frac{1}{50}\)
= 1- \(\frac{1}{50}\)
= \(\frac{49}{50}\)
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
=1/1-1/50
=49/50
đúng đó mik học rồi
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(\Rightarrow A=1-\frac{1}{50}\)
\(\Rightarrow A=\frac{49}{50}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
A=\(1-\frac{1}{50}\)
A=\(\frac{50}{50}-\frac{1}{50}\)
A=\(\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{40}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
A=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
A=1-1/50=49/50
Là 49/50
Mk làm rồi ở bài thi được 10
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\)\(\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{50}{50}-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(1-\frac{1}{50}=\frac{49}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}_{ }\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdot\cdot\cdot\cdot\cdot+\frac{1}{49\cdot50}\)
A=\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{49}-\frac{1}{50}\right)\)
A=\(\left(1-\frac{1}{50}\right)\)
A=\(\frac{49}{50}\)
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
A=\(\frac{1}{1}-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
A=\(\frac{1}{1}-\frac{1}{50}\)
A=\(\frac{49}{50}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow1-\frac{1}{50}=\frac{49}{50}\)
Chúc bạn học tốt !!!
A\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}+\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
Vậy ......
chúc bạn học tốt!
\(A=\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\)\(1-\frac{1}{50}\)
\(A=\)\(\frac{49}{50}\)
Vậy: \(A=\)\(\frac{49}{50}\)
\(A=\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...\)\(+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow1-\frac{1}{50}\)
\(\Rightarrow\frac{49}{50}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\).