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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
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: \(\dfrac{17}{30}=\dfrac{1564}{30\cdot92}\)
\(\dfrac{51}{92}=\dfrac{1530}{30\cdot92}\)
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
câu a ta so sánh số đối của 2 phân số này.nếu ps nào có giá trị tuyệt đối lớn hơn thì nhỏ hơn.
câu b ta nhân cả A và B với 2009 rồi so sánh 2009A với 2009B.ta được A>B
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)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,
\(\dfrac{89}{-13}< 0< \dfrac{1}{123}\\ \Rightarrow\dfrac{89}{-13}< \dfrac{1}{123}\)
Vậy \(\dfrac{89}{-13}< \dfrac{1}{123}\)
b,
\(\dfrac{-13}{15}>\dfrac{-15}{15}=-1=\dfrac{-30}{30}>\dfrac{-31}{30}\)
Vậy \(\dfrac{-13}{15}>\dfrac{-31}{30}\)
c,
\(\dfrac{125}{123}=\dfrac{123}{123}+\dfrac{2}{123}=1+\dfrac{2}{123}\\ \dfrac{99}{97}=\dfrac{97}{97}+\dfrac{2}{97}=1+\dfrac{2}{97}\)
Vì \(\dfrac{2}{97}>\dfrac{2}{123}\Rightarrow1+\dfrac{2}{97}>1+\dfrac{2}{123}\Leftrightarrow\dfrac{99}{97}>\dfrac{125}{123}\)
Vậy \(\dfrac{99}{97}>\dfrac{125}{123}\)
d,
\(\dfrac{125}{126}< \dfrac{126}{126}=1=\dfrac{986}{986}< \dfrac{987}{986}\)
Vậy \(\dfrac{125}{126}< \dfrac{987}{986}\)
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
\(\Rightarrow100-x=0\Rightarrow x=100\)
\(\dfrac{-22}{45}=-\dfrac{2266}{45.103}>-\dfrac{2295}{45.103}=-\dfrac{51}{103}\)
Ta có: \(\dfrac{-22}{45}=\dfrac{-22\cdot103}{45\cdot103}=\dfrac{-2266}{4635}\)
\(\dfrac{-51}{103}=\dfrac{-51\cdot45}{103\cdot45}=\dfrac{-2295}{4635}\)
mà -2266>-2295
nên \(\dfrac{-22}{45}>\dfrac{-51}{103}\)