tìm x biết
x-6/15 - 6/63 - 6/99 - 6/143 - 6/195 = 1/5
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A=6/3.5 + 6/5.7 +....+ 6/15.17
A= 3(1/3-1/5+1/5-1/7+....+1/15-1/17)
A= 3(1/3-1/17)
A= 3.14/51
A=14/17
Vậy....
BÀI 2
60%.x + 0,4.x + x:3= 2
\(\frac{60}{100}\)x + \(\frac{4}{10}\).x + x. \(\frac{1}{3}\)=2
\(\frac{3}{5}\).x + \(\frac{2}{5}\).x + x.\(\frac{1}{3}\)=2
(\(\frac{3}{5}\)+ \(\frac{2}{5}\)+ \(\frac{1}{3}\)) .x =2
\(\frac{4}{3}\).x =2
x = 2: \(\frac{4}{3}\)
x = \(\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
k cho mik nha các bạn
Bài 2 :
60%x + 0.4x + x : 3 = 2
\(x.\left(\frac{60}{100}+\frac{2}{5}+\frac{1}{3}\right)\)= 2
\(x.\frac{4}{3}\)= 2
\(x=2.\frac{3}{4}\)
\(x=1.5\)
2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
=\(1-\frac{1}{15}=\frac{14}{15}\)
\(\Rightarrow S=\frac{7}{15}\)
a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15
A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)
A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)
A=2(1-1/15)
A=1/2.14/15
A=7/15
1/3+1/15+1/35+1/63+1/99+1/143+1/195
=1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+1/11*13+1/13*15
suy ra 2(1/1*3+1/3*5+1/5*7+1/7*9+1/9*11+1/11*13+1/13*15)
=2/1*3+2/3*5+2/5*7+2/7*9+2/9*11+2/11*13+2/13*15
=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15
=1-1/15
=14/15
a=14/15 chia 2=7/15
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\frac{14}{15}\)
\(=\frac{7}{15}\)
Sửa đề chút nhé:
\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)
\(=\left(1+3+5+7+...+2009+2011\right).0\)
\(=0\)
Ý b tham khảo bài bạn nguyen thi thuy linh nhé