1/2x3 + 1/3x4 + 1/4x5 +...+ 1/99x100
CHÚC CÁC BẠN LÀM TỐT
1/1x2+1/2x3+1/3x4+1/4x5+...+1/X nhân ( X + 1 ) = 2017/2018 làm ơn đó
1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 +............. + 1/9x10
bài làm đầy đủ
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+............+1/9+1/10
=1-1/10
=10/10-1/10
=9/10
Bài làm:
\(\frac{1}{1\times2}+\frac{1}{2\times3}\)\(+\frac{1}{3\times4}+\frac{1}{4\times5}\)\(+...\frac{1}{9\times10}\)
\(=\frac{1}{1}-\frac{1}{2}\)\(+\frac{1}{2}-\frac{1}{3}\)\(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}\)\(-\frac{1}{5}\)\(+...\frac{1}{9}-\frac{1}{10}\)
\(=\)\(\frac{1}{1}-\frac{1}{10}\)
\(=\frac{9}{10}\)
(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:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)
\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)