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\(=5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right).\frac{392}{17}\)
\(=5^2\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\left(\frac{1}{8}-\frac{1}{98}\right)\frac{392}{17}\)
\(=25\times\frac{45}{392}\times\frac{392}{17}\)
\(=25\times\frac{45}{17}\)
\(=\frac{1125}{17}\)
Có:
\(\frac{5^3}{8.13}+\frac{5^3}{13.18}+...+\frac{5^3}{93.98}\)
= \(5^2\left(\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{93.98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{93}-\frac{1}{98}\right)\)
=\(25\left(\frac{1}{8}-\frac{1}{98}\right)\)
=\(\frac{1125}{392}\)
=> \(\frac{1125}{392}.3\frac{17}{125}\)
= ...
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+...+\frac{10}{48.53}\)
\(=\frac{10}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+...+\frac{1}{48}-\frac{1}{53}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{53}\right)\)
\(=2.\frac{50}{159}=\frac{100}{159}\)
Ta có: *)A.5=5(1/3.8+1/8.13+...+1/33.38)
=5/3.8+5/8.13+...+5/33.38
=1/3-1/8+1/8-1/13+...+1/33-1/38
=1/3-1/38
=> A=(1/3-1/38).1/5
*)7B=7/3.10+7/10.17+7/17.24+...+7/31.38
=1/3-1/10+1/10-1/17+...+1/31-1/38
=1/3-1/38
=>B=(1/3-1/38).1/7
Do đó a/b=(1/5)/(1/7)=7/5
k mk nha!
\(A=\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}\)
\(A=2\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+\frac{5}{18.23}+\frac{5}{23.28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2.\frac{25}{84}=\frac{25}{42}\)
\(A=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(A=10\left(\frac{1}{3\cdot8}+\frac{1}{8\cdot13}+\frac{1}{13\cdot18}+\frac{1}{18\cdot23}+\frac{1}{23\cdot28}\right)\)
\(A=\frac{10}{5}\left(\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+\frac{1}{18}-\frac{1}{23}+\frac{1}{23}-\frac{1}{28}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{28}\right)\)
\(A=2\cdot\frac{25}{84}\)
\(A=\frac{25}{42}\)
\(B=\frac{10}{3\cdot8}+\frac{10}{8\cdot13}+\frac{10}{13\cdot18}+\frac{10}{18\cdot23}+\frac{10}{23\cdot28}\)
\(B=2\left[\frac{5}{3\cdot8}+\frac{5}{8\cdot13}+\frac{5}{13\cdot18}+\frac{5}{18\cdot23}+\frac{5}{23\cdot28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right]\)
\(B=2\left[\frac{1}{3}-\frac{1}{28}\right]=\frac{25}{42}\)
\(A=1-2+3-4+...+1999-2000+2001\)
\(=\left(1-2\right)+\left(3-4\right)+...+\left(1999-2000\right)+2001\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2001\)
(Từ 1 đến 2000 có 2000 số => có 2000:2=1000 cặp)
\(=\left(-1\right).1000\)
\(=\left(-1000\right)+2001\)
\(=1001\)
(xin lỗi nhe, mik chỉ giúp bạn mỗi câu A thui. Nếu bạn ko k cũng ko sao)
Tìm các số tự nhiên n để phân số A=n+7/n-2 có giá trị là 1 số nguyên
Mọi người giúp mình nha! Cảm ơn mọi người nhé <3
Có \(\frac{5^2}{3.8}+\frac{5^2}{8.13}+\frac{5^2}{13.18}+.....+\frac{5^2}{88.93}\)
=\(\frac{5.5}{3.8}+\frac{5.5}{8.13}+\frac{5.5}{13.18}+...+\frac{5.5}{88.93}\)
= \(5\left(\frac{5}{3.8}+\frac{5}{8.13}+\frac{5}{13.18}+...+\frac{5}{88.93}\right)\)
=\(5\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}+.....+\frac{1}{88}-\frac{1}{93}\right)\)
=\(5\left(\frac{1}{3}-\frac{1}{93}\right)=\frac{50}{31}\)