1/19+9/19.29+9/29.39+...9/1999.2009
Tính nhanh:
\(\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\)
\(\dfrac{1}{19}+\dfrac{9}{19\cdot29}+...+\dfrac{9}{1999\cdot2009}\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{1791}{38171}=\dfrac{200}{2009}\)
Giá trị biểu thức A = 1/19 + 9/19.29 + 9/29.39 +...+ 9/1999.2009
\(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+....+\frac{9}{1999.2009}\)
\(A=\frac{1}{19}+\left(\frac{9}{19.29}+\frac{9}{29.39}+.....+\frac{9}{1999.2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+....+\frac{10}{1999.2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+.....+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
\(A=\frac{1}{19}+\frac{1791}{38171}\)
\(A=\frac{200}{2009}\)
A = 200/2009
đúng 100%, mk thi r` nhưng làm biếng giải
mk đồng ý với bài của Asuka Kurashina
bài của bn ấy đúng 99%
còn 1% là mk ko biết
chúc bn học giỏi!
thanks
tính
E=1/19+9/19.29+9/29.39+...+9/1999.2009
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}.\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}.\left(\frac{1}{19}-\frac{1}{2009}\right)\)
b tự làm nốt nhé
\(\frac{1}{9.19}+\frac{1}{19.29}+\frac{1}{29.39}+...+\frac{1}{1999.2009}\)
\(=\frac{1}{10}\times\left(\frac{10}{9.19}+\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{1}{10}\times\left(\frac{1}{9}-\frac{1}{19}+\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{10}\times\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(=\frac{1}{10}\times\frac{2000}{18081}\)
\(=\frac{200}{18081}\)
_Chúc bạn học tốt_
Ta có:
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+9\left(\frac{1}{19.29}+\frac{1}{29.39}+...+\frac{1}{1999.2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)(Đây là dạng tổng đặc biệt bn nha)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{9}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(=\frac{1}{9}+\frac{9}{10}.\frac{1990}{38171}\)
\(=\frac{1}{9}+\frac{1791}{38171}\)
\(=0,1580...\approx0,16\)
Tính:
1/19+9/19.29+9/29.39+9/39.49+.....+9/1999.2009
ai làm đúng mình tick
Tính:
A=\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{9}-\frac{1}{2009}\right)\)
\(A=\frac{1}{19}+\frac{9}{10}\left(\frac{2000}{18081}\right)\)
\(A=\frac{1}{19}+\frac{200}{2009}\)
\(A=\frac{5809}{38171}\)
MK ko chắc nhé =v ( mấy bước quy đồng lằng nhằng ko làm âu )
Giá trị biểu thức \(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}=?\)
\(A=\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
\(=\frac{1}{19}+\frac{1791}{38171}\)
\(=\frac{200}{2009}\)
Vậy \(A=\frac{200}{2009}\)
Ta có:
\(A=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\) \(\dfrac{9}{1999.2009}\)
\(=\dfrac{1}{19}+\) \(\left(\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\right)\)
\(=\dfrac{1}{19}\) \(+\) \(\dfrac{9}{10}\left(\dfrac{10}{19.29}+\dfrac{10}{29.39}+...+\dfrac{10}{1999.2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)
\(=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)=\dfrac{200}{2009}\)
Vậy \(A=\dfrac{200}{2009}\)
Ta có : \(A=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}.\)\(\Rightarrow A=\dfrac{1}{19}+\left(\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\right)\)\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19.29}+\dfrac{10}{29.39}+..+\dfrac{10}{1999}+\dfrac{10}{2009}\right)\)\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(\Rightarrow A=\dfrac{1}{19}+\dfrac{9}{10}.\dfrac{1990}{38171}\)
\(\Rightarrow A=\dfrac{1}{19}+\dfrac{1791}{38171}\)
\(\Rightarrow A=\dfrac{200}{2009}\)
Vậy \(A=\dfrac{200}{2009}\)
Giá trị của biểu thức \(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
\(\frac{1}{19}+\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\)
= \(\frac{1}{19}+\left(\frac{9}{19.29}+\frac{9}{29.39}+...+\frac{9}{1999.2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}\left(\frac{10}{19.29}+\frac{10}{29.39}+...+\frac{10}{1999.2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+\frac{1}{29}-\frac{1}{39}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)
= \(\frac{1}{19}+\frac{9}{10}.\frac{1990}{38171}\)
= \(\frac{1}{19}+\frac{1791}{38171}\)
= \(\frac{200}{2009}\)
Tính nhanh:
\(B=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\)
\(A=\dfrac{1}{19}+\left(\dfrac{9}{19\cdot29}+\dfrac{9}{29\cdot39}+...+\dfrac{9}{1999\cdot2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{10}{19\cdot29}+\dfrac{10}{29\cdot39}+...+\dfrac{10}{1999\cdot2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{39}+...+\dfrac{1}{1999}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\left(\dfrac{1}{19}-\dfrac{1}{2009}\right)\)
\(A=\dfrac{1}{19}+\dfrac{9}{10}\cdot\dfrac{1990}{38171}\)
\(A=\dfrac{1}{19}+\dfrac{1791}{38171}\)
\(A=\dfrac{200}{2009}\)
B=1/19+(9/19.29+9/29.39+...+9/1999.2009)
B=1/19+9/10+(10/19.29+10/29.39+.....+10/1999.2009
B=1/19+9/10+(1/19-1/29+1/29-1/39+....+1/1999-1/2009)
B=1/19+9/10+(1/19-1/2009)
B=1/19+9/10.1990/38171
B=1/19+1791/38171
B=200/2009
Vậy B= 200/2009