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53x39+47x39-53x21-47x21
=39x ( 53+47) - 21x(53+47)
=39 x100 - 21x100
= 100x(39-21)
=100x18
=1800
2x53x12 +4x6x87 - 3x8x40
= (2x12)x53 + (4x16)x87 - (3x8) x40
= 24x53 + 24x87 + 24x40
= 24( 53+87+40)
= 24x180
=4320
5 x 7 x 77 - 7 x 60 + 49 x 25 -15 x 42
= 5x7x77 - 7x5x12 + 7x7x5x5 - 5x3x7x6
= 35 x77 - 35x12 + 35x35 - 35x18
= 35 x ( 77-12+35-18 )
= 35 x 82 = 2870
a) \(\dfrac{12}{14}=\dfrac{1200}{1400}=\dfrac{1400-200}{1400}=1-\dfrac{200}{1400}\)
\(\dfrac{1212}{1414}=\dfrac{1414-200}{1414}=1-\dfrac{200}{1414}\)
vì \(\dfrac{200}{1414}< \dfrac{200}{1400}\)
Nên \(1-\dfrac{200}{1400}< 1-\dfrac{200}{1414}\)
Vậy \(\dfrac{12}{14}< \dfrac{1212}{1414}\)
Các bài sau tương tự
\(\dfrac{7}{13}\times\dfrac{5}{14}\times\dfrac{39}{15}=\dfrac{7\times5\times3\times13}{13\times7\times2\times5\times3}=\dfrac{1}{2}\)
\(\dfrac{7}{13}\cdot\dfrac{5}{14}\cdot\dfrac{39}{15}=\dfrac{1}{2}\)
\(\dfrac{7}{13}.\dfrac{7}{15}-\dfrac{5}{12}.\dfrac{21}{39}+\dfrac{49}{91}.\dfrac{8}{15}\)
= \(\dfrac{7}{13}.\dfrac{7}{15}-\dfrac{5}{12}.\dfrac{7}{13}+\dfrac{7}{13}.\dfrac{8}{15}\)
= \(\dfrac{7}{13}.\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)
= \(\dfrac{7}{13}.\left(\dfrac{7}{15}+\dfrac{8}{15}-\dfrac{5}{12}\right)\)
= \(\dfrac{7}{13}.\left(1-\dfrac{5}{12}\right)\)
= \(\dfrac{7}{13}.\dfrac{7}{12}=\dfrac{49}{156}\)