A= 42/1.3+42/3.5+42/5.8+....+42/45.47.1-3-5-...-49/8
Mình cần gấp lắm
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a. 4872 - [567 + (124 - 31) x 38]
= 4872 - [567 + 93 x 38]
= 4872 - [567 + 3534]
= 4872 - 4101
= 771
b. 83 : 24 + 513 : 511 x 30
= 32 + 25
= 57
c. 748 x 19 -19 + 19 x 253
= 748 x 19 - 19 x 1 + 19 x 253
= 19 x (748 - 1 + 253)
= 19 x 1000
= 19000
d. 33 x 48 + 34 x 42 + 81 x 42
= 34 x 16 + 34 x 42 + 34 x 42
= 81 x (16 + 42 + 42)
= 81 x 100
= 8100
~Study well~
#Seok_Jin
a.4872-[567+(124-31).38]
a=4872-[567+93.38]
a=4872-[567+3534]
a=4872-4101
a=771
Câu a thôi nhé , các câu còn lại cậu tự làm nhé!
\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)
\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)
\(\dfrac{\dfrac{7}{13}+\dfrac{7}{14}-\dfrac{7}{15}}{\dfrac{8}{13}+\dfrac{8}{14}-\dfrac{8}{15}}-\dfrac{\dfrac{5}{11}-\dfrac{5}{13}+\dfrac{5}{15}}{\dfrac{8}{11}-\dfrac{8}{13}+\dfrac{8}{15}}\)
\(=\dfrac{7\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}\right)}{8\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}\right)}-\dfrac{5\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{15}\right)}{8\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{15}\right)}\)
\(=\dfrac{7}{8}-\dfrac{5}{8}\)
\(=\dfrac{2}{8}=\dfrac{1}{4}\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
68 x 35 =2380
175 x 42 = 7350
1023 x 29 = 29667
nhớ k cho mình nhé bạn
Thực hiện phép nhân và phép chia ở cả 2 vế và so sánh.
Em điền được kết quả như sau:
a) 7 × 5 > 7 × 4 | 7 × 2 = 2 × 7 | 7 × 8 < 7 × 9 |
b) 42 : 7 < 42 : 6 | 21 : 7 = 6 : 2 | 56 : 7 > 49 : 7 |
\(\left(49\cdot42-47\cdot42\right):42\)
\(=\left[\left(49-47\right)\cdot42\right]:42\)
\(=2\cdot42:42\)
\(=2\)
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)