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a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
\(\frac{16.17-5}{16.16+11}\)
= \(\frac{16.\left(16+1\right)-5}{16.16}\)
= \(\frac{16.16+16-5}{16.16+11}\)
= \(\frac{16.16+11}{16.16+11}\)
= \(1\)
\(\frac{16.17-5}{16.16+11}=\frac{17-5}{16+11}=\frac{8}{27}\)
Hok tốt ~~
K chắc đâu
# DanLinh
\(\frac{1}{5}+\frac{4}{10}+...+\frac{81}{45}=\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+...+\frac{9}{5}=\frac{1+2+3+...+9}{5}=\frac{45}{5}=9\)
1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
=1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
=45/5 = 9
\(\left(a\right)\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{36+15}{32}=\frac{51}{32}\)
\(\left(b\right)4+\frac{35}{45}=\frac{4}{1}+\frac{35}{45}=\frac{180}{45}+\frac{35}{45}=\frac{180+35}{45}=\frac{43}{9}\)
\(\left(c\right)\frac{11}{4}-\frac{15}{16}=\frac{44}{16}-\frac{15}{16}=\frac{44-15}{16}=\frac{29}{16}\)
\(\left(d\right)3-\frac{13}{9}=\frac{3}{1}-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{27-13}{9}=\frac{14}{9}\)
\(\left(e\right)\frac{5}{6}-\frac{5}{8}=\frac{20}{24}-\frac{15}{24}=\frac{20-15}{24}=\frac{5}{24}\)
\(\left(g\right)\frac{196}{64}-2=\frac{196}{64}-\frac{2}{1}=\frac{196}{64}-\frac{128}{64}=\frac{196-128}{64}=\frac{17}{16}\)
Mình đảm bảo đúng ! Chúc bạn học tốt!
a) \(\frac{9}{8}+\frac{15}{32}=\frac{36}{32}+\frac{15}{32}=\frac{51}{32}\)
b) \(4+\frac{35}{45}=4+\frac{7}{9}=\frac{36}{9}+\frac{7}{9}=\frac{43}{9}\)
c)\(\frac{11}{4}+\frac{15}{16}=\frac{44}{16}+\frac{15}{16}=\frac{59}{16}\)
d )\(3-\frac{13}{9}=\frac{27}{9}-\frac{13}{9}=\frac{14}{9}\)
e ) \(\frac{5}{6}+\frac{5}{8}=\frac{40}{48}+\frac{30}{48}\)\(=\frac{70}{48}\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{41}+1\)\(-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(A=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=9-\left[1\div\left(1\times2\right)+1\div\left(2\times3\right)+1\div\left(3\times4\right)+1\div\left(4\times5\right)\right]\)\(+1\div\left(5\times6\right)+1\div\left(6\times7\right)+1\div\left(7\times8\right)+1\div\left(8\times9\right)\)\(+1\div\left(9\times10\right)\)]
\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)\(+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\))
\(A=9-\left(1-\frac{1}{10}\right)\)
\(A=9-\frac{9}{10}\)
\(A=\frac{81}{10}\)
\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(A=9-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{10-9}{10.9}\right)\)
\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9-\frac{9}{10}=\frac{81}{10}\)
=(1-1/2)+(1-1/6)+(1-1/12)+...+(1-1/90)
=9-(1/2 + 5/6 + 1/12 + ... + 1/90)
=9-(1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10)
=9-(1-1/10)
=9-9/10=81/10
=(1-1/2)+(1-1/6)+(1-1/12)+.......+(1-1/90)
= 9 - (1/2 +5/6 +1/12+.......+1/90)
= 9- (1-1/2 + 1/2 - 1/3+1/3 -1/4 +....... +1/9-1/10)
=9-(1-1/10)
=9-9/10=81/10
=(1-1/2)+(1-1/6)+(1-1/12)+.......+(1-1/90)
= 9 - (1/2 +5/6 +1/12+.......+1/90)
= 9- (1-1/2 + 1/2 - 1/3+1/3 -1/4 +....... +1/9-1/10)
=9-(1-1/10)
=9-9/10=81/10
a) \(\frac{16.17-5}{16.16+11}=\frac{16.16+16-5}{16.16+11}=\frac{16.16+11}{16.16+11}=1\)1
b ) \(\frac{5}{30}+\frac{15}{90}+\frac{25}{150}+\frac{35}{210}+\frac{45}{270}=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\)
\(=\frac{1}{6}.5=\frac{5}{6}\)
a) (16x17-5)/(16x16+11)=(16x16-5+16)/(16x16+11)=1
b) (1x5)/(6x5)+(3x5)/(18x5)+...+(9x5)/(54x5)
=1/6+1/6+1/6+1/6+1/6
=5/6