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a. \(\dfrac{-18}{91}\) và \(\dfrac{-23}{114}\)
\(\dfrac{18}{91}< \dfrac{18}{90}=\dfrac{1}{5}=\dfrac{23}{115}< \dfrac{23}{114}\)
\(\Rightarrow\dfrac{-18}{91}>\dfrac{-23}{114}\)
b. \(\dfrac{-22}{35}\) và \(\dfrac{-103}{177}\)
\(\dfrac{22}{35}=\dfrac{110}{175}>\dfrac{103}{175}>\dfrac{103}{177}\)
\(\Rightarrow\dfrac{-22}{35}< \dfrac{-103}{177}\)
a: \(\dfrac{-18}{91}>\dfrac{-18}{90}=\dfrac{-1}{5}\)
\(\dfrac{-23}{114}< \dfrac{-23}{115}=\dfrac{-1}{5}\)
Do đó: \(\dfrac{-18}{91}>\dfrac{-23}{114}\)
b: \(\dfrac{-22}{35}=\dfrac{-3894}{6195}\)
\(\dfrac{-103}{177}=\dfrac{-103\cdot35}{177\cdot35}=\dfrac{-3605}{6195}\)
mà -3894<-3605
nên -22/35<-103/177
a: \(\dfrac{119}{117}=1+\dfrac{2}{117}\)
\(\dfrac{117}{115}=1+\dfrac{2}{115}\)
mà 2/117<2/115
nên \(\dfrac{119}{117}< \dfrac{117}{115}\)
hay \(-\dfrac{119}{117}>-\dfrac{117}{115}\)
b: \(\dfrac{-22}{35}=\dfrac{-22\cdot177}{35\cdot177}=-\dfrac{3894}{6195}\)
\(\dfrac{-103}{177}=\dfrac{-103\cdot35}{177\cdot35}=\dfrac{-3605}{6195}\)
mà -3894<-3605
nên -22/35<-103/177
a)Vì \(\dfrac{21}{52}>0\); \(\dfrac{-213}{523}< 0\)
\(\Rightarrow\)\(\dfrac{21}{52}>\dfrac{-213}{523}\)
b)Ta có :\(\dfrac{22}{35}< 1\) ;\(\dfrac{103}{17}>1\)
\(\Rightarrow\)\(\dfrac{22}{35}< \dfrac{103}{17}\)
c)Ta có :\(\dfrac{-2525}{4949}=\dfrac{-25}{49}\) ;\(\dfrac{-131313}{373737}=\dfrac{-13}{37}\)
Lại có :\(\dfrac{25}{49}>\dfrac{25}{50}=\dfrac{1}{2}\) hay \(\dfrac{-25}{49}< \dfrac{-25}{50}=\dfrac{-1}{2}\)(1)
và \(\dfrac{13}{37}< \dfrac{13}{26}=\dfrac{1}{2}\) hay \(\dfrac{-13}{37}>\dfrac{-13}{26}=\dfrac{-1}{2}\) (2)
Từ (1) và (2) \(\Rightarrow\)\(\dfrac{-2525}{4949}< \dfrac{-131313}{373737}\)
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: \(\dfrac{17}{30}=\dfrac{1564}{30\cdot92}\)
\(\dfrac{51}{92}=\dfrac{1530}{30\cdot92}\)
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
Bài 1 :
a, \(-1\dfrac{2}{3}\)= \(\dfrac{-5}{3}\)
Dựa vào tính chất của Tỉ lệ thức :
Ta có : \(\dfrac{x}{y}=\dfrac{-5}{3}\rightarrow\dfrac{x}{-5}=\dfrac{y}{3}\)
Dựa vào tính chất của dãy tỉ số = nhau
Ta có : \(\dfrac{x}{-5}=\dfrac{y}{3}=\dfrac{x+y}{\left(-5\right)+3}=\dfrac{18}{-2}=-9\)
\(\rightarrow\dfrac{x}{-5}=-9\rightarrow x=\left(-5\right).\left(-9\right)\Rightarrow x=45\\ \rightarrow\dfrac{y}{3}=-9\rightarrow y=3.\left(-9\right)\Rightarrow y=-27\)b,
Ta có :
( x + 4 ) . 7 = ( y + 7 ) . 4
\(\rightarrow\) 7x + 28 = 4y + 28
\(\rightarrow\) 7x = 4y
Vì 7x = 4y
\(\Rightarrow\) x = 22 / ( 4 + 7 ) . 7 = 14
\(\Rightarrow\) y = 22 - 14 = 8
Đợi mk lm câu 2 nha
a. \(\dfrac{-18}{91}\) và \(\dfrac{-23}{114}\) ( mẫu chung : 10374 )
Quy đồng : \(\dfrac{-18}{91}=\dfrac{-2052}{10374}\) ; \(\dfrac{-23}{114}=\dfrac{-2093}{10374}\)
Vì \(\dfrac{-2052}{10374}>\dfrac{-2093}{10374}\Rightarrow\dfrac{-18}{91}>\dfrac{-23}{114}\)
Vậy...
b. \(\dfrac{-22}{35}\) và \(\dfrac{-103}{177}\) ( MC = 6195 )
Quy đồng : \(\dfrac{-22}{35}=\dfrac{-3894}{6195};\dfrac{-103}{177}=\dfrac{-3605}{6195}\)
Vì \(\dfrac{-3894}{6195}< \dfrac{-3605}{6195}\Rightarrow\dfrac{-22}{35}< \dfrac{-103}{177}\)
Vậy...
c. \(\dfrac{-22}{45}\) và \(\dfrac{-17}{33}\)(MC=495)
Quy đồng : \(\dfrac{-22}{45}=\dfrac{-242}{495};\dfrac{-17}{33}=\dfrac{-255}{495}\)
Vì \(\dfrac{-242}{495}>\dfrac{-255}{495}\Rightarrow\dfrac{-22}{45}>\dfrac{-17}{33}\)
Vậy