1 tính
A=1/1.2+1/2.3+1/3.4…+1/97.98+1/98.99+1/99.100
Chú thích : dấu chấm là dấu nhân
Ai lm xong thi mình tink cho 3 cái lun ok
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\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=1-\dfrac{1}{99}\)
\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)
Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)
Suy ra \(A>B\).
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)
1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.
=1/1-1/2022
=2021/2022
Đặt A=1.98+2.97+3.96+...+96.3+97.2+98.1
B=1.2+2,3+3.4+...+96.97+97.98+98.99
Ta có: A=1+(1+2)+...+(1+2+3+...+97+98)
=\(\dfrac{1.2}{2}+\dfrac{2.3}{2}+...+\dfrac{98.99}{3}\)
=\(\dfrac{1.2+2.3+3.4+4.5+...+98.99}{2}\)=\(\dfrac{B}{2}\)
=>E=\(\dfrac{B}{2}\):2=\(\dfrac{1}{2}\)
A = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}...-\dfrac{1}{97.98}\)
A= 1-\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{97.98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)
A= 1- \(\left(\dfrac{1}{1}-\dfrac{1}{98}\right)\)
A=1- 1 + \(\dfrac{1}{98}\)
A= \(\dfrac{1}{98}\)
Lời giải:
$1-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{97.98}$
$1-A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{98-97}{97.98}$
$1-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{97}-\frac{1}{98}$
$=1-\frac{1}{98}$
$\Rightarrow A=\frac{1}{98}$
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{!}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{1024}+\frac{1}{2048}\)
\(\Rightarrow\)\(2C=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}+\frac{1}{1024}\)
\(\Rightarrow\)\(2C-C=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Leftrightarrow\)\(C=1-\frac{1}{2048}=\frac{2047}{2048}\)
...
= 1/2-1/3+1/3-1/4+...+ 1/19-1/20
= 1/2-1/20
=9/20
có phải như thế này ko bn
\(A=\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{19}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}\)
A = \(\frac{9}{20}\)
\(B=\frac{1}{99.100}-\frac{1}{98.99}-\frac{1}{97.98}-.....-\frac{1}{1.2}=-\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{99.100}\right)\)
\(B=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=-\left(1-\frac{1}{100}\right)\)
B = \(-\frac{99}{100}\)
C=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+....+\(\frac{1}{97.98}\)+\(\frac{1}{98.99}\)
Câu hỏi của lương hiếu - Toán lớp 6 - Học toán với OnlineMath
Làm như link trên nhưng bỏ hạng tử \(\frac{1}{99.100}\)đi
Bước cuối: \(1-\frac{1}{99}=\frac{98}{99}\)
Ta có : A = 1/1.2 + 1/2.3 + .... + 1/98.99 + 1/99.100 .
=> A = 1 - 1/2 + 1/2 - 1/3 + .... + 1/98 - 1/99 + 1/99 - 1/100 .
=> A = 1 - 1/100 .
=> A = 99/100 .
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}\)