\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) . Tính
Bài: tính
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
= 1/2-1/3+ 1/3 -1/4 +... +1/99-1/100
=1/2-1/100
=50/100 - 1/100= 49/100
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Tham khảo nha !!!
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Tính A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{99.100}=?\)
mk bít lm cách lớp 5, vừa học
Cần ko bn
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=?\)
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
1/1 . 2 + 1/ 2 . 3 + 1/ 3 . 4 + ... + 1/99 . 100
= 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 100/100 + -1/100
= 99/100
#Hoq chắc _ Baccanngon
tính giá trị của biểu thức
\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)
\(A=1+\frac{99}{100}=\frac{199}{100}\)
=1+1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1+1/2+1/2-1/100
=199/100
A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A=1+1/1-1/2+1/2-1/3+1/4+...+1/98-1/99+1/99-1/100
A-1=1-1/100
A-1=99/100
A=99/100+1
A=199/100
Vậy A=199/100
Tính giá trị của biểu thức
\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(=\frac{199}{100}\)
Gọi biểu thức là A
A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1-1/2+1/2-1/3+1/3-1/4+...+/198-1/99+1/99-1/100
A-1=1-1/100
A-1=99/100
A=99/100+1
A=199/100
cái này bấm máy tính cũng ra nek
Tính
E=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)có lời giải nhé😘😘😘
E = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
E = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
E = \(\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
E = \(\frac{1}{2}-\frac{1}{100}\)
E = \(\frac{49}{100}\)
Tính :
\(\frac{1}{1.3}+\)\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
Mk cần gấp :v
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Sửa lại đề bài nha bạn:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
Chúc em học tốt nhé!
Xin phép sửa đề 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/99.100
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
tính
\(\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(A=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot...\cdot1-\frac{2}{99\cdot100}\)
\(2A=1-\left(\frac{1}{2\cdot3}\cdot\frac{1}{3\cdot4}\cdot\frac{1}{4\cdot5}\cdot...\cdot\frac{1}{99\cdot100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}-\frac{1}{5}\cdot...\cdot\frac{1}{99}\cdot\frac{1}{100}\right)\)
\(2A=1-\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(2A=1-\frac{49}{100}\)
\(2A=\frac{51}{100}\)
\(A=\frac{51}{100}:2\)
\(A=\frac{51}{200}\)
Tính
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5..100}\)
\(=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
\(=2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)...2\left(\frac{1}{2}-\frac{2}{99.100}\right)\)
\(=2^{89}.\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)\)
= \(\frac{2^{98}.4851}{100}\)