Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn Kiên giải đúng nhưng chưa rõ nên mình giải lại.
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{202}{201}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{\left(x+1\right)}\right)=\frac{202}{201}\)
\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{202}{201}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+1\right)}=\frac{202}{201}:2=\frac{202}{402}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{202}{402}=-\frac{1}{402}=\frac{-1}{402}=\frac{1}{-402}\)
\(\Rightarrow\frac{1}{x+1}=\hept{\begin{cases}\frac{-1}{402}\\\frac{1}{-402}\end{cases}}\Rightarrow x+1=\hept{\begin{cases}402\\-402\end{cases}}\Rightarrow\hept{\begin{cases}x=402-1\\x=\left(-402\right)-1\end{cases}}\Rightarrow x=\hept{\begin{cases}401\\-403\end{cases}}\)
\(\Rightarrow A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{202}{201}\)\(\Rightarrow A=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{202}{201}\)
\(\Rightarrow A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{202}{201}\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{202}{201}\)
\(\Rightarrow A=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{202}{201}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{202}{402}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{202}{402}=\frac{-1}{402}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{-402}\)
\(\Rightarrow x+1=-402\)
\(\Rightarrow x=-403\)
\(A=\frac{15\times3^{11}+4\times27^4}{9^7}\)
\(A=\frac{15\times177147+4\times531441}{4782969}\)
\(A=\frac{2657205+2125764}{4782969}\)
\(A=\frac{47829969}{47829969}=1\)
(1-1/3).(1-1/6).(1-1/10)...(1-1/780) (ko có dấu "...")
Giải: (1-1/3).(1-1/6).(1-1/10)...(1-1/780)
= 2/3 . 5/6....779/780
= 4/6 . 10/12.....1558/1560
= 1.4 . 2.5 .... 38.41/ 2.3 . 3. 4. .....39.40
= ( 1.2.3....38).(4.5....41)/(2.3.4....39)(3...
triệt tiêu xong còn 41/39.3= 41/117
ĐS = 41/117
k cho mh nha bạn
a) 25/42 - 20/63 = 5/18
b) ( 9/50 - 13/75 - 1/6 ) : ( -1/2 )2 = -16/25
c) ( 2/15 - 2/65 - 4/39 ) : ( 1/2 - 1/3 )2 = 0
kick và kết bạn với mình nha
a) Ta có: \(\dfrac{7\cdot25}{14\cdot10}\)
\(=\dfrac{7\cdot5\cdot5}{7\cdot2\cdot2\cdot5}\)
\(=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{24\cdot15-14\cdot9}{36\cdot12}\)
\(=\dfrac{9\cdot8\cdot5-14\cdot9}{36\cdot12}\)
\(=\dfrac{9\cdot\left(8\cdot5-14\right)}{9\cdot4\cdot12}\)
\(=\dfrac{40-14}{4\cdot12}\)
\(=\dfrac{13}{24}\)
=2*(1/6+1/12+1/20+...+1/380)
=2*(1/2*3+1/3*4+1/4*5+...+1/19*20)
=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/19-1/20)
=2*(1/2-1/20)
=2*(10/20-1/20)
=2*9/20
=18/20
=9/10
A/2=(1/3+1/6+1/10+1/15+...+1/171+1/190)/2
=1/6+1/12+1/20+1/30+...+1/342+1/380
=1/2.3+1/3.4+1/4.5+1/5.6+....+1/18.19+1/19.20
=1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +....+ 1/18 - 1/19 + 1/19 - 1/20
=1/2 - 1/20 = 10/20 - 1/20 = 9/20
=> A = 9/20 .2 = 18/20=9/10