LN
2
K
Khách
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LT
0
17 tháng 4 2016
D = -1-1/3-1/6-1/10-...-1/1225
Suy ra : D/2=-1/2-1/6-1/12-....-1/2450
Mà 1/2=1/(1.2)=1-1/2; 1/6=1/(2.3)=1/2-1/3;...1/2450=1/(49.50)=...
D/2= -(1-1/2)-(-1/2-1/3)-...-(1/49-1/50)
D/2= -1+1/2-1/2+1/3-....-1/49+1/50
D/2= -1+1/50=-49/50
D=(-49/50).2=-98/50
k nha
11 tháng 2 2017
ta có
1/2 P=1/2(1-1/10-1/15-1/3-1/28-1/6-1/21)
=1/2-(1/6+1/12+1/20+1/30+1/42+1/56)
=1/2-(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8)
=1/2-(1/2-1/8)
=1/8
suy ra P=1/4
11 tháng 2 2017
ta có
1/2 P=1/2(1-1/10-1/15-1/3-1/28-1/6-1/21)
=1/2-(1/6+1/12+1/20+1/30+1/42+1/56)
=1/2-(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8)
=1/2-(1/2-1/8)
=1/8
suy ra P=1/4
H9
HT.Phong (9A5)
CTVHS
5 tháng 11 2023
Ta có:
\(\left(x-1\right)^2+\left(y+2\right)^2=0\)
Do: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)
Mặt khác: \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Thay vào B ta có:
\(B=2\cdot1^5-5\cdot\left(-2\right)^3+4=2\cdot1-5\cdot-8+4=2+40+4=46\)
TN
1
DT
2
VG
12 tháng 12 2023
A = -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)
= -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))
Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)
Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))
= 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))
= 2×\(\dfrac{24}{50}\)
= \(\dfrac{24}{25}\)
Thay B vào A ta có :
A = -1-\(\dfrac{24}{25}\)
=> A = \(\dfrac{-49}{25}\)
Cho mik một tick nhé thankss
NH
0
NH
0
\(B=-1-\frac{1}{3}-\frac{1}{6}-...-\frac{1}{1225}\)
\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)
\(B=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
\(B=-2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\right)\)
\(B=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(B=-2\left(1-\frac{1}{50}\right)\)
\(B=-2\cdot\frac{49}{50}\)
\(B=-\frac{49}{25}\)