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10,42 : (21,34 - \(\frac{1}{2}\))+\(\frac{2}{3}\).75%
=10,42 : 20,84 +\(\frac{2}{3}\).75%
=\(\frac{1}{2}\)+\(\frac{2}{3}\).75:100
=\(\frac{1}{2}\)+50:100
=\(\frac{1}{2}\)+\(\frac{1}{2}\)=1
\(=0.75-\dfrac{7}{3}-0.75+9\cdot\left(-\dfrac{1}{9}\right)=-\dfrac{7}{3}-1=-\dfrac{10}{3}\)
\(\left(\dfrac{-2}{3}\right).0,75+\dfrac{1}{\dfrac{2}{3}}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)\)
\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}:\left(\dfrac{-8}{18}\right)+\left(\dfrac{-9}{18}\right)\)
\(=\dfrac{-1}{2}+\dfrac{5}{3}.\dfrac{-18}{8}+\dfrac{-9}{18}\)
\(=\dfrac{-1}{2}+\dfrac{-15}{4}+\dfrac{-9}{18}\)
\(=\dfrac{-72}{144}+\dfrac{-540}{144}+\dfrac{-72}{144}\)
\(=\dfrac{-684}{144}\)
\(=\dfrac{-19}{4}.\)
0,75 = \(\dfrac{3}{4}\)
Ta có: \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... +\(\dfrac{1}{2000.2001}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2000}\) - \(\dfrac{1}{2001}\).
<=> \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{1}{2}\) - \(\dfrac{1}{2001}\).
Vì \(\dfrac{1}{2}\) < \(\dfrac{3}{4}\) nên \(\dfrac{1}{2}\) - \(\dfrac{1}{2001}\) < \(\dfrac{3}{4}\).
Vậy \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{2000^2}\) < \(\dfrac{3}{4}\).
\(D=\frac{13}{8}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-25\%.\frac{1}{2}\)
\(D=\frac{13}{8}+\frac{1}{8}:\left(\frac{3}{4}-\frac{1}{2}\right)-\frac{1}{4}X\frac{1}{2}\)
\(D=\frac{13}{8}+\frac{1}{8}:\frac{1}{4}-\frac{1}{4}X\frac{1}{2}\)
\(D=\frac{13}{8}+\frac{4}{8}-\frac{1}{8}\)
\(D=\frac{17}{8}-\frac{1}{8}\)
\(D=2\)
18/3+1/8÷(3/4-1/2)-1/4.1/2
18/3+1/8÷-1/4-1/4.1/2
1/8.-4/1-1/4.1/2
-4/8-1/8
-5/8
kq cua mk la 1
10,42 : (21,34 - 1/2) + 2/3 * 0,75
= 10,42 : 20,84 + 0,5
= 0,5 + 0,5
=1