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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
7 tháng 7 2023
Bài 1
a) \(5\times72\times10\times2=\left(5\times2\times10\right)\times72=100\times72=7200\)
b) \(40\times125=5\times\left(8\times125\right)=5\times1000=5000\)
c) \(16\times6\times25=4\times4\times6\times25=\left(4\times6\right)\times\left(4\times25\right)=24\times100=2400\) Bài 2:
a) \(24\times57+43\times24=24\times\left(57+43\right)=24\times100=2400\)
b) \(12\times19+80\times12+12=12\times\left(19+80+1\right)=12\times100=1200\)
c) \(\left(36\times15\times169\right)\div\left(5\times18\times13\right)\)
\(=\left(18\times2\times3\times5\times13\times13\right)\div\left(5\times18\times13\right)\)
\(=\left(2\times3\times13\right)\times\left(18\times5\times13\right)\div\left(5\times18\times13\right)\)
\(=2\times3\times13\)
\(=78\)
d) \(\left(44\times52\times60\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times11\times4\times13\times4\times15\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times4\times4\right)\times\left(11\times13\times15\right)\div\left(11\times13\times15\right)\)
\(=4\times4\times4\)
\(=64\)
Bài 3:
a) \(x-280\div35=5\times54\)
\(x-8=270\)
\(x=270+8\)
\(x=278\)
b) \(\left(x-280\right)\div35=54\div4\)
\(\left(x-280\right)\div35=\dfrac{27}{2}\)
\(x-280=\dfrac{27}{2}\times35\)
\(x-280=\dfrac{945}{2}\)
\(x=\dfrac{945}{2}+280\)
\(x=\dfrac{1505}{2}\)
c) \(\left(x-128+20\right)\div192=0\)
\(x-128+20=0\times192\)
\(x-128+20=0\)
\(x-128=0-20\)
\(x-128=-20\)
\(x=-20+128\)
\(x=108\)
d) \(4\times\left(x+200\right)=460+85\times4\)
\(4\times\left(x+200\right)=460+340\)
\(4\times\left(x+200\right)=800\)
\(x+200=800\div4\)
\(x+200=200\)
\(x=200-200\)
\(x=0\)
Bài 4:
a) \(\dfrac{7}{12}-\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
b) \(\dfrac{8}{11}+\dfrac{19}{11}=\dfrac{27}{11}\)
c) \(\dfrac{3}{8}+\dfrac{5}{12}=\dfrac{9}{24}+\dfrac{10}{24}=\dfrac{19}{24}\)
d) \(\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{9}{12}+\dfrac{7}{12}=\dfrac{16}{12}=\dfrac{4}{3}\)
Bài 5:
a) \(x-\dfrac{6}{7}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}+\dfrac{6}{7}\)
\(x=\dfrac{47}{14}\)
b) \(\dfrac{12}{7}\div x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\dfrac{12}{7}\div x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\dfrac{12}{7}\div x=\dfrac{11}{15}\)
\(x=\dfrac{12}{7}\div\dfrac{11}{15}\)
\(x=\dfrac{180}{77}\)
VT
20 tháng 8 2017
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
làm giống như trên
\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)
\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)
\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)
P/S: . là nhân nha
NT
6
QA
11 tháng 8 2016
a) 1+2+3+...+99+100
giải
Từ 1 đến 100 có 100 số.Như vậy,số cặp số là:
100:2=50(cặp)
Mỗi cặp số có tổng bằng:
1+100(2+99)(3+98)...=11
Kết quả của phép tính là:
101x50=5050
Đáp số:5050
31 tháng 8 2023
a) 10; 13; 18; 26; 36; 52...
c) 0; 1; 4; 9; 16; 25...
m) 1; 4; 9; 16; 25; 36; 49; 64...
p) 1; 3; 9; 27; 81; 243...