1/1x2 + 1/2x3 + 1/3x4 +.........................+ 1/15x16
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}........+\frac{1}{15x16}=?\)
x là nhân nhé. Nhanh và đúng mình tick
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.16}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}=\frac{15}{16}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{15x16}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{15}-\frac{1}{16}\)
\(=1-\frac{1}{16}\)
\(=\frac{15}{16}\)
1/1x2+1/2x3+1/3x4+1/24x25
1/1x2+ 1/2x3+1/3x4+1/24x25
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
[x+1/2x3] + [x+1/3x4] + ...+[x+ 1/15x16]=39/16
đề bài tìm x
hêu mi
\(\left(x+\frac{1}{2\cdot3}\right)+\left(x+\frac{1}{3\cdot4}\right)+....+\left(x+\frac{1}{15\cdot16}\right)=\frac{39}{16}\)
\(\Leftrightarrow\left(x+x+x+....+x\right)+\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{15\cdot16}\right)=\frac{39}{16}\)
\(\Leftrightarrow14x+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{15}-\frac{1}{16}\right)=\frac{39}{16}\)
\(\Leftrightarrow14x+\left(\frac{1}{2}-\frac{1}{16}\right)=\frac{39}{16}\)
\(\Leftrightarrow14x+\frac{7}{16}=\frac{39}{16}\)
\(\Leftrightarrow14x=2\)
\(\Leftrightarrow x=\frac{1}{7}\)
[X+1/2x3] + [X+1/3x4] + ...+[X+ 1/15x16]=39/16
tìm X
(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