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\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{6}{12}=\dfrac{193}{1066}\)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
\(A=17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=\left(17\dfrac{2}{31}-6\dfrac{2}{31}\right)-\dfrac{15}{17}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
\(B=\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{12}=36\dfrac{363}{533}-36\dfrac{1}{2}=\dfrac{193}{1066}\) (Casio :>)
\(C=27\dfrac{51}{59}-\left(7\dfrac{51}{59}-\dfrac{1}{3}\right)=27\dfrac{51}{59}-7\dfrac{51}{59}+\dfrac{1}{3}\)
\(=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 .
= ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) .
= 2 + ( 20/21 + 7/21 ) .
= 2 + 9/7 .
= 23/7 .
b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) .
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41 .
= 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) .
= 1/3 - 1 + 1 .
= 1/3 .
c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 .
= [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 .
= [ 11/4 . ( - 1 ) ] . 8/33 .
= -11/4 . 8/33 .
= -2/3 .
d, 38/45 - ( 8/45 - 17/51 - 3/11 ) .
= 38/45 - 8/45 + 17/51 + 3/11 .
= 2/3 + 17/51 + 3/11 .
= 374/561 + 187/561 + 153/561 .
= 14/11 .
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
a) \(\frac{17}{30}>\frac{51}{92}\)
b) \(\frac{-45}{47}>\frac{31}{-30}\)
c) \(\frac{22}{67}< \frac{51}{152}\)
d) \(-\frac{17}{39}< -\frac{17}{41};\frac{18}{-39}< -\frac{17}{41}\)
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
a) 17x3/30x3 và 51 /92
=51/90 > 51/92
b) -3x3 /5x3 và -9/23
-9/15> -9/23
c)-15x10101/23x10101 và -151515/232323
=-151515/232323=-151515/232323
a: \(\dfrac{17}{30}=\dfrac{1564}{30\cdot92}\)
\(\dfrac{51}{92}=\dfrac{1530}{30\cdot92}\)
mà 1564>1530
nên 17/30>51/92
b: \(\dfrac{-45}{47}>-1>-\dfrac{31}{30}\)
c: \(\dfrac{67}{22}=3+\dfrac{1}{22}\)
\(\dfrac{152}{51}=3+\dfrac{1}{51}\)
mà 1/22>1/51
nên 67/22>152/51
=>22/67<51/152
d: 17/39>17/41
nên -17/39<-17/41
=>-18/39<-17/39<-17/41