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\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(4x+\frac{15}{16}=\frac{23}{16}\)
\(4x=\frac{1}{2}\)
\(x=\frac{1}{8}\)
Vậy \(x=\frac{1}{8}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Rightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=\frac{23}{16}\)
\(\Rightarrow5x+\frac{15}{32}=\frac{23}{16}\)
\(\Rightarrow5x=\frac{23}{16}-\frac{15}{32}\)
\(\Rightarrow5x=\frac{31}{32}\)
\(\Rightarrow x=\frac{31}{32}.\frac{1}{5}=\frac{31}{160}\)
\(a,\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\)
\(b,\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)
\(=\frac{1\times2\times3}{2\times3\times4}=\frac{1}{4}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{100}\right).200x=4036\)
\(\Leftrightarrow\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{99}{100}.200x=4036\)
\(\Leftrightarrow\frac{1.2.3...99}{2.3.4....100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow\frac{1}{100}.200x=4036\)
\(\Leftrightarrow2x=4036\)
\(\Leftrightarrow x=4036:2=2018\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{100}\right)\times200\times x=4036\)
=> \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}\times200\times x=4036\)
=> \(\frac{1\times2\times...\times99}{2\times3\times...\times100}\times200\times x=4036\)
\(\Rightarrow\frac{1}{100}\times200\times x=4036\)
\(\Rightarrow2\times x=4036\)
=> x = 2018
\(A=\frac{1+2+...+2014}{2015}\)
\(A=\frac{\frac{2014\cdot2015}{2}}{2015}\)
\(A=\frac{1007\cdot2015}{2015}\)
\(A=1007\)
câu này ở trong Violympic nên mình nói luôn đáp án là 1/8
= 1 + 2 + 3 + ... + 2014( 1007 số hạng) / 2015 = ( 2014 + 1 ) . 1007 / 2015 = 2015 . 1007 / 2015 = 1007
1-(1-(1/2 × 3/5 + 1/5 × 1/2 + 2/10 × 1/2 +1/2)
=1-(1-(3/10 + 1/10 + 1/10 + 5/10)
=1-(1-(3+1+1+5/10)
=1-(1- 10/10)
=1-(1-1)
=1-0
=1
\(1-\left[1-\left(\frac{1}{2}\times\frac{3}{5}+\frac{1}{5}\times\frac{1}{2}+\frac{2}{10}\times\frac{1}{2}+\frac{1}{2}\right)\right]\)
\(=1-\left[1-\left\{\frac{1}{2}\times\left(\frac{3}{5}+\frac{1}{5}+\frac{2}{10}+\frac{1}{2}\right)\right\}\right]\)
\(=1-\left[1-\left\{\frac{1}{2}\times\left(\frac{6}{10}+\frac{2}{10}+\frac{2}{10}+\frac{5}{10}\right)\right\}\right]\)
\(=1-\left[1-\left\{\frac{1}{2}\times\left(\frac{6+2+2+5}{10}\right)\right\}\right]\)
\(=1-\left[1-\left\{\frac{1}{2}\times\frac{3}{2}\right\}\right]\)
\(=1-\left[1-\frac{3}{4}\right]\)
\(=1-\frac{1}{4}\)
\(=\frac{3}{4}\)
[ 19/20 x 3/4 + 1/20 x 3/4 ] x [ 1/2 x 3/4 -2/4 x 3/4 ]
=[ 3/4 x [19/20 + 1/20 ] x [3/4 x [2/4 - 1/2]
=[ 3/4 x 1 ] x [3/4 x 0 ]
=0
\(\left(\frac{19}{20}.\frac{3}{4}+\frac{1}{20}.\frac{3}{4}\right).\left(\frac{1}{2}.\frac{3}{4}-\frac{2}{4}.\frac{3}{4}\right)\)
\(=\left(\frac{19}{20}.\frac{3}{4}+\frac{1}{20}.\frac{3}{4}\right).\left(\frac{1}{2}.\frac{3}{4}-\frac{1}{2}.\frac{3}{4}\right)\)
\(=\left(\frac{19}{20}.\frac{3}{4}+\frac{1}{20}.\frac{3}{4}\right).0\)
\(=0\)