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\(\frac{6.4+6.7}{6.5+12}=\frac{6.4+6.7}{6.5+6.2}=\frac{6.\left(4+7\right)}{6.\left(5+2\right)}=\frac{11}{7}\)
\(\frac{6.4+6.7}{6.5+12}=\frac{4+1.7}{1.5+2}=\frac{4+7}{5+2}=\frac{11}{7}\)
\(\frac{6\cdot4+6\cdot7}{6\cdot5+12}\)
\(=\frac{6\left(4+7\right)}{6\cdot5+6\cdot2}\)
\(=\frac{6\left(4+7\right)}{6\left(5+2\right)}\)
\(=\frac{11}{7}\)
a)2.3.5.13/26.35
=2.3.5.13/2.13.5.7
=3/7
b)18.6-18/-36.(-35)
=18.6-18/-(36.25)=18(6-1)/-(36.35)
=18.5/-(36.35)=2.3.3.5/-(2.2.3.3.5.7)
=-1/14
c)6.4+6.7/6.5+12=6.4+6.7/6.5+6.2
=6(4+7)/6(5+2)
=11/10
d)25.9-25.17/-8.80-8.10=25(9-17)/8(-80-10)
=25.(-8)/8(-90)
=5/18
e) 34.5-36/34.13-34
=34(5-32)/34(13-1)
=5-32/13-1=-1/3
a ) \(\frac{-360}{450}\)
TA có : \(\frac{-360}{450}=\frac{-4}{5}\)
b ) \(\frac{-260}{1500}\)
Ta có : \(\frac{-260}{1500}\)= \(\frac{-13}{75}\)
\(P=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)
\(P=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)
\(P=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(P=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(P=\frac{1}{2}.\frac{50}{101}\)
\(P=\frac{25}{101}\)
\(P=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{6.7}+...+\frac{1}{198.101}\)
\(P=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{6.14}+...+\frac{1}{198.202}\right)\)
\(P=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{6}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(P=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(P=\frac{1}{2}.\frac{50}{101}\)
\(P=\frac{25}{101}\)
a/\(\frac{3939-101}{3.2929+505}=\frac{39.101-101}{8787+505}=\frac{101.\left(39-1\right)}{87.101+5.101}=\frac{101.38}{101.\left(87+5\right)}=\frac{38}{92}\)
\(=\frac{38}{92}\)
b/\(\frac{6.4+6.7}{6.5+12}=\frac{4+1.7}{1.5+2}=\frac{4+7}{5+2}=\frac{11}{7}\)
\(\frac{6^5+6^6.7}{6^3+6^6.4}=\frac{6^5.\left(1+6.7\right)}{6^3.\left(1+6^3.4\right)}\)
\(=\frac{6^2.43}{865}=\frac{36.43}{865}\)
\(=\frac{1568}{865}\)