tính tống sau bằng cách hợp lý:
A= \(\dfrac{1}{2x3}\)\(+\)\(\dfrac{1}{3x4}+\dfrac{1}{4x5}+\dfrac{1}{5x6}+\dfrac{1}{6x7}\)
Tính bằng cách thuận tiện nhất:
\(\dfrac{1}{2x3}\) + \(\dfrac{1}{3x4}\) + \(\dfrac{1}{4x5}\) + \(\dfrac{1}{5x6}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{1}{2}-\dfrac{1}{6}\)
\(=\dfrac{1}{3}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{1}{3}\)
Tính tổng sau thành hợp lý: A = 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7 + 1/7x8 ?( ghi cả phép tính nha )
`A=1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+1/(6xx7)+1/(7xx8)`
`=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8`
`=1/2-1/8`
`=4/8-1/8`
`=3/8`
Vậy `A=3/8`
tính bằng cách thuận tiện: 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7
=1/2-1/3+1/3-1/4+...+1/6-1/7
=1/2-1/7
=5/14
Câu 10 (1,0 điểm)
Cho S = \(\dfrac{1}{2x3}+\dfrac{1}{4x5}+\dfrac{1}{6x7}+...+\dfrac{1}{2020x2021}+\dfrac{1}{2022x2023}\)
So sánh S với \(\dfrac{1011}{2023}\)
so sánh A và B biết
A=\(\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{99x100}\)
B=\(\dfrac{1}{1x3}+\dfrac{1}{3x5}+\dfrac{1}{5x7}+...+\dfrac{1}{97x99}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{99}=\dfrac{49}{99}>\dfrac{49}{100}=A\)
\(\dfrac{1}{1x2}\)+\(\dfrac{1}{2x3}\)+\(\dfrac{1}{3x4}\)+......+\(\dfrac{1}{9x10}\)
Tính bằng cách nhanh nhất
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
#kễnh
tính bằng cách hợp lý nhất : 1/2x3 +1/3x4 +1/4x5 + 1/5x6
1/2x3 +1/3x4 +1/4x5 + 1/5x6 = 7/18
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}\)
\(=\frac{1}{3}\)
tính nhanh
1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7
1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/6x7 + 1/20x21
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{21-2}{42}=\dfrac{19}{42}\)
Lời giải:
Gọi biểu thức số 1 là A và số 2 là B
\(A=\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
B tương tự A:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)
Tính nhanh:
A,(\(\dfrac{8}{15}\)+\(\dfrac{14}{23}\))-(\(\dfrac{5}{15}\)-\(\dfrac{9}{23}\))
B,\(\dfrac{1}{1x2}\)+\(\dfrac{1}{2x3}\)+\(\dfrac{1}{3x4}\)+\(\dfrac{1}{4x5}\)+\(\dfrac{1}{5x6}\)
A, \(\left(\dfrac{8}{15}+\dfrac{14}{23}\right)-\left(\dfrac{5}{15}-\dfrac{9}{23}\right)\)
\(=\dfrac{8}{15}+\dfrac{14}{23}-\dfrac{5}{15}+\dfrac{9}{23}\)
\(=\left(\dfrac{8}{15}-\dfrac{5}{15}\right)+\left(\dfrac{14}{23}+\dfrac{9}{23}\right)\)
\(=\dfrac{3}{15}+1\)
\(=1\dfrac{1}{5}\)
B, \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=1-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
a) \(=\dfrac{8}{15}+\dfrac{14}{23}-\dfrac{5}{15}+\dfrac{9}{23}\)
\(=\dfrac{8}{15}-\dfrac{5}{15}+\dfrac{14}{23}+\dfrac{9}{23}\)
\(=\dfrac{1}{5}+1\)
\(=\dfrac{6}{5}\)
b)
b) \(=\dfrac{1}{2x}\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2x}\left(2+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}\right)\)
\(=\dfrac{1}{2x}[2\left(1+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)]\)
\(=\dfrac{1}{2x}[2\left(1+\dfrac{1}{3}\right)]\)
\(=\dfrac{1}{2x}\left(2.\dfrac{4}{3}\right)\)
\(=\dfrac{4}{3x}\)