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\(\left[\frac{139}{84}-\frac{577}{924}-\frac{98}{205}\right].\frac{2197}{27}\)
\(\left[\frac{34}{33}-\frac{98}{205}\right]\frac{2197}{27}\)
\(\frac{3736}{6765}.\frac{2197}{27}=44,93713285\)
P=(1-1/1+2).(1-1/1+2+3)....(1-1/1+2+3+4+...+2011)
=[1-1/(2+1).2:2].[1-1/(3+1).3:2].....[1-1/(2011+1).2011:2]
=(1-2/2.3).(1-2/3.4)...(1-2/2011.2012)
=4/2.3.10/3.4....4046130/2011.2012
=1.4/2.3 .2.5/3.4 ....2010.2013/2011.2012
=1.2....2010/2.3...2011 .4.5....2013/3.4....2012
=1/2011.2013/3
=671/2011
1/ Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1\)
Dấu "=" xảy ra khi x=y=1
Máy mình bị lỗi nên ko nhìn được các bài tiếp theo
Chúc bạn học tốt :)
Ta có : x+y=2 => x=2-y. Thay vào bt ta đc : xy= (2-y).y = 2y -y^2
Vì y^2 >= 0 =>2y-y^2 nhỏ hơn hoặc bằng 0
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)
\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)
\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}.\frac{2.\left(2+1\right)}{2}+\frac{1}{3}.\frac{3.\left(3+1\right)}{2}+...+\frac{1}{16}.\frac{16.\left(16+1\right)}{2}\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=\frac{\left(17-2+1\right).\left(17+2\right)}{2}:2\)
\(=76\)
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\frac{1}{2}\left[\frac{\left(2+1\right)2}{2}\right]+\frac{1}{3}\left[\frac{\left(3+1\right)3}{3}\right]+...+\frac{1}{16}\left[\frac{\left(16+1\right)16}{2}\right]\)
\(=1+\frac{2+1}{2}+\frac{3+1}{2}+...+\frac{16+1}{2}\)
\(=\frac{2+2+1+3+1+...+16+1}{2}\)
\(=\frac{\left(1+1+1+..15cs.+1\right)+\left(2+3+...+16\right)+2}{2}\)
\(=\frac{15+135+2}{2}\)
\(=\frac{152}{2}\)\(=76\)
=(-1/2) : (-2/3) :( -3/4) :...: (-49/50)
= -1/2 . (-3/2) . (-4/3) . ... . (-50/49)
= -1/2.(-1/2) . (-50)
= - 1/100
mik ko bik viết kiểu của bn nên khó nhìn, thông cảm nha