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a) \(\frac{30}{51}\)\(-\)\(\frac{20}{52}\)\(+\)\(\frac{14}{34}-\frac{56}{91}-2\)
\(=\frac{10}{17}-\frac{5}{13}+\frac{7}{17}-\frac{8}{13}-2\)
= (\(\frac{10}{17}\)\(+\frac{7}{17}\)) \(-\)(\(\frac{5}{13}+\frac{8}{13}\)) \(-2\)
= \(1-1-2\)
=\(0-2\)
= -2
k cho mik nha
a) 93. 85+ 17. 45-170 = 93. 17. 5+ 17.9.5-17.5.2= 17.5.( 93+ 9-2)= 85. 100 = 8500
b) \(\left(\frac{-13}{31}:\frac{5}{17}+\frac{-18}{31}:\frac{5}{17}\right).\frac{25}{34}\)
\(=\left(\frac{-13}{31}.\frac{17}{5}+\frac{-18}{31}.\frac{17}{5}\right).\frac{25}{34}\)
\(=\left(\frac{-13}{31}+\frac{-18}{31}\right).\frac{17}{5}.\frac{25}{34}\)
\(=\left(-1\right).\frac{5}{2}\)
\(=\frac{-5}{2}\)
CHÚC BN HỌC TỐT!!
a, 98.85 + 17.45 - 170 = 116.3 - 170 = -53.7
b, (-13/31 : 5/17 + -18/31 : 5/17) x 25/34 = 17/5 x ( -13/31 + -18/31) x 25/34
=17/5 x -1 x 25/34 = -5/2 = -2.5
\(\frac{22}{45}\)< \(\frac{51}{101}\)
\(\frac{23}{48}\)< \(\frac{47}{92}\)
\(\frac{34}{43}\)< \(\frac{35}{42}\)
xét A và B có :
\(\frac{42}{47}\)<\(\frac{42}{45}\) (1)
theo tính chất bắc cầu ta có ;
\(\frac{37}{51}\)+\(\frac{14}{51}\)=1 ; \(\frac{29}{37}\)+\(\frac{8}{37}\)=1
\(\frac{31}{35}\)+\(\frac{4}{35}\)=1 ; \(\frac{49}{63}\)+\(\frac{14}{63}\)=1
Mà \(\frac{14}{51}\)>\(\frac{14}{63}\)=> \(\frac{37}{51}\)< \(\frac{49}{63}\)(2)
ta lại có : \(\frac{4}{35}\)=\(\frac{8}{70}\)( nhân cả tử và mẫu vs 2 )
mà \(\frac{8}{70}\)<\(\frac{8}{37}\)nên \(\frac{4}{35}\)<\(\frac{8}{37}\)=>\(\frac{29}{37}< \frac{31}{35}\)(3)
Từ (1) ; (2);(3)=>\(\frac{42}{47}+\frac{37}{51}+\frac{29}{37}< \frac{42}{45}+\frac{49}{63}+\frac{31}{35}\)
\(60!=1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot59\cdot60=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2\cdot4\cdot6\cdot...\cdot58\cdot60\)
\(=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\cdot1\cdot2\cdot3\cdot...\cdot30=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\times30!\)
\(\Rightarrow1\cdot3\cdot5\cdot...\cdot59=\frac{60!}{30!\times2^{30}}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}\)đpcm.
\(\frac{31}{2}\cdot\frac{32}{2}\cdot...\cdot\frac{60}{2}\cdot2\cdot4\cdot...\cdot58\cdot60\)
=31.32.33.34...60.1.2.3.4.5...29.30
=1.2.3.4.5.6.7.8.9.10...60
1.3.5.7...59.2.4.6.8...60
=1.2.3.4.5.6...60
Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}=1\cdot3\cdot5\cdot...\cdot59\)