SO SÁNH A VÀ B
a ) A=\(\dfrac{3}{8^3}+\dfrac{7}{^{ }8^4}\) B =\(\dfrac{7}{8^3}+\dfrac{3}{8^4}\)
b ) A=\(\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}\) B=\(\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\)
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d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
Ta có: \(\dfrac{20}{39}>\dfrac{20}{41}>\dfrac{18}{41}\left(1\right)\)
\(\dfrac{22}{27}>\dfrac{22}{29}\left(2\right)\)
\(\dfrac{18}{43}=1-\dfrac{25}{43};\dfrac{14}{39}=1-\dfrac{25}{39}\)
Vì \(\dfrac{25}{43}< \dfrac{25}{39}\Rightarrow1-\dfrac{25}{43}>1-\dfrac{25}{39}\Rightarrow\dfrac{18}{43}>\dfrac{14}{39}\left(3\right)\)
Từ \(\left(1\right);\left(2\right);\left(3\right)\) ta suy ra : A>B
Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
\(a,\dfrac{11}{22}-\dfrac{3}{16}.\dfrac{8}{18}+\dfrac{1}{18}=\dfrac{1}{2}-\dfrac{1}{12}+\dfrac{1}{18}=\dfrac{18}{36}-\dfrac{3}{36}+\dfrac{2}{36}=\dfrac{17}{36}\)
\(b,\dfrac{5}{7}+\dfrac{3}{7}:4-\dfrac{8}{9}.\dfrac{-3}{4}=\dfrac{5}{7}+\dfrac{3}{7}.\dfrac{1}{4}-\dfrac{-2}{3}=\dfrac{5}{7}+\dfrac{3}{28}+\dfrac{2}{3}=\dfrac{60}{84}+\dfrac{9}{84}+\dfrac{56}{84}=\dfrac{125}{84}\)
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
a, Ta có:
A= \(\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
B= \(\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\) nên A < B.
b, Ta có:
\(\dfrac{20}{39}>\dfrac{14}{39}\)
\(\dfrac{22}{27}>\dfrac{22}{29}\)
\(\dfrac{18}{43}< \dfrac{18}{41}\)
\(\Rightarrow\)\(\dfrac{20}{39}+\dfrac{22}{27}+\dfrac{18}{43}>\dfrac{14}{39}+\dfrac{22}{29}+\dfrac{18}{41}\)
Hay A > B