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Bài này dễ ợt
\(\left(1+1+1\right)!=6\)
\(2+2+2=6\)
\(3\times3-3=6\)
\(\sqrt{4}+\sqrt{4}+\sqrt{4}=6\)
\(5+5:5=6\)
\(6-6+6=6\)
\(7-7:7=6\)
\(\sqrt{\left(8+8:8\right)}!=6\)
\(\sqrt{9}\times\sqrt{9}-\sqrt{9}=6\)
\(\sqrt{\left(10-10:10\right)}!=6\)
1a)\(\frac{5}{3}\)=\(\frac{5x4}{3x4}\)=\(\frac{20}{12}\); \(\frac{1}{4}\)=\(\frac{1x3}{4x3}\)=\(\frac{3}{12}\)
b)\(\frac{3}{8}\)=\(\frac{3x3}{8x3}\)=\(\frac{9}{24}\); \(\frac{7}{24}\)
c)\(\frac{1}{2}\)=\(\frac{1x15}{2x15}\)=\(\frac{15}{30}\); \(\frac{2}{3}\)=\(\frac{2x10}{3x10}\)=\(\frac{20}{30}\); \(\frac{3}{5}\)=\(\frac{3x6}{5x6}\)=\(\frac{18}{30}\)
2a)\(\frac{11}{8}\)>\(\frac{11}{9}\)
b)\(\frac{4}{9}\)<\(\frac{3}{5}\)
c)\(\frac{6}{5}\)>\(\frac{5}{6}\)
1.3.77−1+3.7.99−3+7.9.1313−7+9.13.1515−9+\frac{19-13}{13.15.19}+13.15.1919−13
=\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}=1.31−3.71+3.71−7.91+7.91−9.131+9.131−13.151+13.151−15.191
=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31−15.191=28595−2851=28594
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)b,=61.(1.3.76+3.7.96+7.9.136+9.13.156+13.15.196)
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)c,=81.(1.2.31+2.3.41+3.4.51+...+48.49.501)
=\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)=161.(1.2.32+2.3.42+3.4.52+...+48.49.502)
=\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)=161.(1.2.33−1+2.3.44−2+3.4.55−3+...+48.49.5050−48)
=\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)=161.(1.21−2.31+2.31−3.41+3.41−4.51+...+48.491−49.501)
=\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}=161.(21−24501)=161.(24501225−24501)=4900153
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)d,=75.(1.5.87+5.8.127+8.12.157+...+33.36.407)
=\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)=75.(1.5.88−1+5.8.1212−5+8.12.1515−8+...+33.36.4040−33)
=\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)=75.(1.51−5.81+5.81−8.121+8.121−12.151+...+33.361−36.401)
=\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}=75.(51−14401)=75.(1440288−14401)=28841
P/S: . là nhân nha
Tổng của các số trong mỗi ô nhỏ là 26
6 + 8 + 7 + 5 = 26
8 + 8 + 6 + 4 = 26
10 + 9 + 4 + 3 = 26
Vậy số cần tìm là: 10
hay lam ho viet cac so co ba chu so 1,2,3 ma moi chu so chi viet mot lantrong mot so va phan nguyen co mot chu so
a; Cách một:
\(\dfrac{2}{9}\) = \(\dfrac{2\times2}{9\times2}\) = \(\dfrac{4}{18}\) < \(\dfrac{4}{10}\) Vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
\(\dfrac{4}{9}\) = \(\dfrac{4\times3}{9\times3}\) = \(\dfrac{12}{27}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times2}{10\times2}\) = \(\dfrac{12}{20}\)
Vì \(\dfrac{12}{27}\) < \(\dfrac{12}{20}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{12}{20}\)
\(\dfrac{3}{8}\) = \(\dfrac{3\times4}{8\times4}\) = \(\dfrac{12}{24}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times3}{7\times3}\) = \(\dfrac{12}{21}\)
Vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
\(\dfrac{5}{9}\) = \(\dfrac{5\times7}{9\times7}\) = \(\dfrac{35}{63}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times5}{10\times5}\) = \(\dfrac{35}{50}\)
Vì \(\dfrac{35}{63}\) < \(\dfrac{35}{50}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
Cách hai:
a; \(\dfrac{2}{9}\) = \(\dfrac{2\times10}{9\times10}\) = \(\dfrac{20}{90}\); \(\dfrac{4}{10}\) = \(\dfrac{4\times9}{10\times9}\) = \(\dfrac{36}{90}\)
Vì \(\dfrac{20}{90}\) < \(\dfrac{36}{90}\) vậy \(\dfrac{2}{9}\) < \(\dfrac{4}{10}\)
b; \(\dfrac{4}{9}\) = \(\dfrac{4\times10}{9\times10}\) = \(\dfrac{40}{90}\); \(\dfrac{6}{10}\) = \(\dfrac{6\times9}{10\times9}\) = \(\dfrac{54}{90}\)
Vì \(\dfrac{40}{90}\) < \(\dfrac{54}{90}\) vậy \(\dfrac{4}{9}\) < \(\dfrac{6}{10}\)
c; \(\dfrac{3}{8}\) = \(\dfrac{3\times7}{8\times7}\) = \(\dfrac{21}{56}\); \(\dfrac{4}{7}\) = \(\dfrac{4\times8}{7\times8}\) = \(\dfrac{32}{56}\)
Vì \(\dfrac{21}{56}\) < \(\dfrac{32}{56}\) vậy \(\dfrac{3}{8}\) < \(\dfrac{4}{7}\)
d; \(\dfrac{5}{9}\) = \(\dfrac{5\times10}{9\times10}\) = \(\dfrac{50}{90}\); \(\dfrac{7}{10}\) = \(\dfrac{7\times9}{10\times9}\) = \(\dfrac{63}{90}\)
Vì \(\dfrac{50}{90}\) < \(\dfrac{63}{90}\) vậy \(\dfrac{5}{9}\) < \(\dfrac{7}{10}\)
4 1/2 < 4 3/4.
2 4/5 < 3 1/4.
7 2/9 > 5 2/9.
13 5/6 < 13 6/7.
Ban k cho minh voi nha.
\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)
\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)
\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)
\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)
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
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