2 ,63 x 99 + 2,63
=
=
=
K
Khách
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ND
Nguyễn Đức Trí
VIP
15 tháng 7 2023
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
OY
18 tháng 7 2021
Sửa đề
\(\dfrac{3}{35}+\dfrac{3}{63}+\dfrac{3}{99}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)
\(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)
\(\dfrac{1}{x+2}=\dfrac{-9}{35}\)
\(x+2=35\)
\(x=35-2\)
\(x=33\)
\(-17x=-17-\left(-34\right)\)
\(-17x=-17\)
\(x=-1\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
Bạn coi lại đề: $\frac{3}{69}$ không nằm trong các số hạng có quy luật.
YM
3
3 tháng 7 2018
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+....+\frac{2}{575}+\frac{2}{675}\)
\(=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{23\times25}+\frac{2}{25\times27}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{23}-\frac{1}{25}+\frac{1}{25}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
Vậy ....
3 tháng 7 2018
\(\frac{2}{15}+\frac{2}{35}+\frac{2}{69}+........+\frac{2}{575}+\frac{2}{675}\)
\(=\frac{2}{3\times5}+\frac{2}{5\times7}+........+\frac{2}{23\times25}+\frac{2}{25\times27}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{23}-\frac{1}{25}+\frac{1}{25}-\frac{1}{27}\)
\(=\frac{1}{3}-\frac{1}{27}\)
\(=\frac{8}{27}\)
U
1
5 tháng 3 2022
\(\Leftrightarrow\dfrac{3}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{x+2}=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{5}\)
=>x+2=5
hay x=3
KC
6
WH
19 tháng 5 2018
\(17,8-\left(14,904:3\cdot y-1,084\right)-2,63=13,77\)
\(\Leftrightarrow4,968\cdot y-1,084=17,8-\left(13,77+2,63\right)\)
\(\Leftrightarrow4,968\cdot y-1,084=17,8-16,4\)
\(\Leftrightarrow4,968\cdot y=17,8-16,4+1,084\)
\(\Leftrightarrow4,968\cdot y=2,484\)
\(\Leftrightarrow y=2,484:4,968\)
\(\Leftrightarrow y=0,5\)
19 tháng 5 2018
\(17,8-\left(14,904:3.y-1,084\right)-2,63=13,77\)
\(\Leftrightarrow17,8-\left(4,968.y-1,084\right)=13,77+2,63\)
\(\Leftrightarrow17,8-\left(4,968.y-1,084\right)=16,4\)
\(\Leftrightarrow4,968.y-1,084=17,8-16,4\)
\(\Leftrightarrow4,968.y-1,084=1,4\)
\(\Leftrightarrow4,968.y=1,4+1,084\)
\(\Leftrightarrow4,968.y=2,484\)
\(\Leftrightarrow y=2,484:4,968\)
\(\Leftrightarrow y=0,5\)
Vậy y = 0,5
_Chúc bạn học tốt_
30 tháng 8 2022
=>x(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+99/x=-3/7
=>8/9x+99/x=-3/7
\(\Leftrightarrow\dfrac{8x}{9}+\dfrac{99}{x}=\dfrac{-3}{7}\)
\(\Leftrightarrow\dfrac{8x^2+99\cdot9}{9x}=\dfrac{-3}{7}\)
\(\Leftrightarrow-56x^2-6237=27x\)
hay \(x\in\varnothing\)
NH
30 tháng 3 2019
\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)
Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(A=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{9.11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
Thay A vào biểu thức
\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)
\(\Rightarrow x=\frac{22}{15}\)
P/s: Có thể tính sai :(
30 tháng 3 2019
\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)
Trước tiên mình tính dãy có dấu ngoặc đã
Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)
\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)
Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :
\(\frac{5}{11}\times x=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)
\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)
\(\Leftrightarrow x=\frac{22}{15}\)
Vậy \(x=\frac{22}{15}\)
2,63 x 99 + 2,63
= 2,63 x 99 + 2,63 x 1
= 2,63 x ( 99 + 1 )
= 2,63 x 100
= 263
Học tốt ! ^^
2,63 x 99 + 2,63
= 2,63 x (99 + 1)
= 2,63 x 100
= 263