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\(A=\frac{3469-54}{6938-108}\)
\(=\frac{3415}{6830}\)
\(=\frac{3415}{6830}=\frac{3415:3415}{6830:3415}=\frac{1}{2}=\frac{3}{6}\)
\(B=\frac{2468-89}{3720-147}\)
\(=\frac{2370}{3555}\)
\(=\frac{2370}{3555}=\frac{2370:1185}{3555:1185}=\frac{2}{3}=\frac{4}{6}\)
1/a có số số hạng là:(2016-6):2+1=1006(số)
Tổng A là:(2016+6)x1006:2=1017066
Cauu B sai
c có số số hạng là:(150-3):3+1=50(số)
Tổng C là: (150+3)x50:2=3825
2/ Số cuối cùng là:72+79x2=230
a,18 . 76 + 24 . 18 - 150 = 18 . ( 76 + 24 ) - 150= 18 . 100 - 150 = 1800 - 150=1650
b, 147 + 230 + 53 + 70 = ( 147 + 53 ) + ( 230 + 70) = 200 + 300=500
c, 7 . 85 + 27 . 7 - 7 . 12=(85 + 27 -12 ) . 7=100 . 7=700
Dấu . là phép nhân
1) 18 . 76 + 24 . 18 -150
= 18( 76 + 24 ) - 150
= 18 . 100 -150
= 1800 - 150
= 1650
2) 147+230 +53 + 70
= ( 147 + 53 ) + ( 230 + 70 )
= 200 + 200
= 400
3) 7 × 85 + 27 × 7 - 7 ×12
= 7( 85 + 27 - 12 )
= 7 . 100
= 700
\(A=\frac{12}{3.5}+\frac{12}{5.7}+...+\frac{12}{2013.2015}\)
\(2A=\frac{24}{3.5}+\frac{24}{5.7}+...+\frac{24}{2013.2015}\)
\(2A=\frac{24}{3}-\frac{24}{5}+\frac{24}{5}-\frac{24}{7}+...+\frac{24}{2013}-\frac{24}{2015}\)
\(2A=8-\frac{24}{2015}\)
\(2A=\frac{8}{1}-\frac{24}{2015}\)
\(2A=\frac{16120}{2015}-\frac{24}{2015}\)
\(2A=\frac{16096}{2015}\)
\(=>A=\frac{16096}{2015}:2\)
\(=>A=\frac{16096}{4030}\)
d) 135(-12 + 247) - 147(135 - 12)= -12.135 + 135.147 - 147.135 + 12.147 = (-12.135 + 12.147) + (135.147 - 147.135) = 12(-135 + 147) + 0 = -12.12 = -144
Bài 1:
a) [3.(-2) - (-8)].(-7) - (-2).(-5) = -7(-6 + 8) - 2.5= -7.2 - 2.5 = -2(7 + 5) = -2.12 = -24
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{241864704+1209323520}{2176782336-362797056}=\frac{4}{5}\)
a) \(2\frac{3}{13}-\frac{5}{9}-\left(\frac{3}{13}+\frac{4}{9}\right)\)
= \(\frac{29}{13}-\frac{5}{9}-\left(\frac{3}{13}+\frac{4}{9}\right)\)
= \(\left(\frac{29}{13}-\frac{3}{13}\right)-\left(\frac{5}{9}+\frac{4}{9}\right)\)
= \(2-1\)
= \(1\)
b) \(17\frac{4}{16}+\frac{3}{4}-\left(2\frac{3}{12}+75\%\right)\)
= \(\frac{69}{4}+\frac{3}{4}-\left(\frac{27}{12}+\frac{3}{4}\right)\)
= \(\left(\frac{69}{4}+\frac{3}{4}\right)-\left(\frac{27}{12}+\frac{3}{4}\right)\)
= \(18-3\)
= \(15\)
c) \(\frac{6}{5.7}+\frac{6}{7.9}+\frac{6}{9.11}+....+\frac{6}{101.103}+\frac{6}{103.106}\)
= \(3.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+....+\frac{2}{101.103}+\frac{2}{103.106}\right)\)
= \(3.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{101}-\frac{1}{103}+\frac{1}{103}-\frac{1}{106}\right)\)
= \(3.\left(\frac{1}{5}-\frac{1}{106}\right)\)
= \(3.\frac{101}{530}\)
= \(\frac{303}{530}\)