Tính
-1 - 1/2 - 1/4 -1/8 -... - 1/1024
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a. \(\frac{1}{5}+\frac{3}{4}+\frac{1}{10}\)
= \(\frac{4}{20}+\frac{15}{20}+\frac{2}{20}\)
= \(\frac{21}{20}\)
b. \(\frac{5}{6}-\frac{1}{3}+\frac{1}{6}\)
= \(\frac{5}{6}-\frac{2}{6}+\frac{1}{6}\)
= \(\frac{4}{6}=\frac{2}{3}\)
c. \(\frac{3}{8}-\frac{10}{2}:\frac{4}{5}\)
= \(\frac{3}{8}-\frac{50}{8}\)
= \(\frac{-47}{8}\)
a) \(\frac{1}{5}+\frac{3}{4}+\frac{1}{10}\)
= \(\frac{4+15+2}{20}\)
= \(\frac{21}{20}\)
b) \(\frac{5}{6}-\frac{1}{3}+\frac{1}{6}\)
= \(\frac{5-2+1}{6}\)
= \(\frac{4}{6}\)
c) \(\frac{3}{8}-\frac{10}{2}:\frac{4}{5}\)
= \(\frac{3}{8}-\frac{25}{4}\)
= \(-\frac{47}{8}\)
\(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)+ \(\frac{1}{64}\)+ \(\frac{1}{128}\)= \(\frac{123}{234}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{b-a}{5-4}=4\)
Do đó: a=16; b=20
Gọi số bi của An và Chi ll là a,b(viên;a,b∈N*)
Áp dụng tc dtsnb:
\(\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{b-a}{5-4}=4\\ \Leftrightarrow\left\{{}\begin{matrix}a=16\\b=20\end{matrix}\right.\)
Vậy ...
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\)
A\(\times\) 2 = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 1 - \(\dfrac{1}{128}\)
A\(\times\)(2-1) = \(\dfrac{128-1}{128}\)
A = \(\dfrac{127}{128}\)
Gọi \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là B
\(B=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2\cdot B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(2\cdot B-B=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{32}+\dfrac{1}{64}-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(B=1+\left(\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+.....+\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(B=1+0-\dfrac{1}{128}\)
\(B=1-\dfrac{1}{128}\)
\(B=\dfrac{128}{128}-\dfrac{1}{128}\)
\(B=\dfrac{127}{128}\)
1) x-2x-3x-4x-5x=-13x thay x=-8 => -13x=(-8).(-13)=104
2)(x+1)-2(x+1)-3(x+1)-4(x+1)-5(x+1)=-13(x+1) thay x=-8 => -13(x+1)=(-13).(-7)-91