1/12+1/2x3+1/3x4+1/4x5
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
haizzz đáng tiếc tôi muốn ns là: ko bao f và đừng mong chờ OK
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
Lên Qanda mà hỏi
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
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:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)
1/2x3+1/3x4+1/4x5+....+1/99x100=?
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
ai lam dc mik tick cho