\(\frac{6.4+6.7}{6.5+12}=\frac{4+1.7}{1.5+2}=\frac{4+7}{5+2}=\frac{11}{7}\)

\(\frac{6\cdot4+6\cdot7}{6\cdot5+12}\)

\(=\frac{6\left(4+7\right)}{6\cdot5+6\cdot2}\)

\(=\frac{6\left(4+7\right)}{6\left(5+2\right)}\)

\(=\frac{11}{7}\)

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 : \(\frac{-260}{1500}\)= \(\frac{-13}{75}\)

ta có : \(A=6.7+6.7^2+6.7^3+...+6.7^{100}\)

\(\Rightarrow7A=7.\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)

\(7A=6.7^2+6.7^3+6.7^4+...+6.7^{101}\)

\(\Rightarrow7A-A=6A=\left(6.7^2+6.7^3+6.7^4+...+6.7^{101}\right)-\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)

\(6A=6.7^{101}-6.7=6\left(7^{101}-7\right)\Leftrightarrow A=7^{101}-7\)

vậy \(A=7^{101}-7\)

ta có : \(B=6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)

\(\Rightarrow5B=5\left(6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\right)\)

\(5B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}\)

\(\Rightarrow5B+B=6B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}+6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)

\(6B=6.5-6.5^{101}=6.\left(5-5^{101}\right)\Leftrightarrow B=5-5^{101}\)

vậy \(B=5-5^{101}\)

\(A=6\cdot7+6\cdot7^2+6\cdot7^3+...+6\cdot7^{100}\\ =6\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot7+\left(7-1\right)\cdot7^2+\left(7-1\right)\cdot7^3+...+\left(7-1\right)\cdot7^{100}\\ =7^2-7+7^3-7^2+7^4-7^3+...+7^{101}-7^{100}\\ =7^{101}-7=7\cdot\left(7^{100}-1\right)\)

\(B=6\cdot5-6\cdot5^2+6\cdot5^3-...+6\cdot5^{99}-6\cdot5^{100}\\ =6\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\ =\left(5+1\right)\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\=\left(5+1\right)\cdot5-\left(5+1\right)\cdot5^2+\left(5+1\right)\cdot5^3-...+\left(5+1\right)\cdot5^{99}-5^{100}\\ =5^2+5-5^3-5^2+5^4+5^3+...+5^{100}+5^{99}-5^{101}-5^{100}\\ =5-5^{101}\\ =5\cdot\left(1-5^{100}\right)\)

\(P=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

\(P=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{10.14}+...+\frac{1}{198.202}\right)\)

\(P=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(P=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(P=\frac{1}{2}.\frac{50}{101}\)

\(P=\frac{25}{101}\)

\(P=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{6.7}+...+\frac{1}{198.101}\)

\(P=2.\left(\frac{1}{2.6}+\frac{1}{6.10}+\frac{1}{6.14}+...+\frac{1}{198.202}\right)\)

\(P=2.\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{6}-\frac{1}{14}+...+\frac{1}{198}-\frac{1}{202}\right)\)

\(P=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)

\(P=\frac{1}{2}.\frac{50}{101}\)

\(P=\frac{25}{101}\)

mấy bạn ơi phân số thứ 3 mình ghi sai nó vậy nè \(\frac{1}{10.7}\)

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/\(\frac{6.4+6.7}{6.5+12}=\frac{4+1.7}{1.5+2}=\frac{4+7}{5+2}=\frac{11}{7}\)

\(\frac{6.4+6.7}{6.5+12}\)\(=\frac{6.4+6.7}{6.5+6.2}\)\(=\frac{6.\left(4+7\right)}{6.\left(5+2\right)}\)\(=\frac{6.11}{6.7}\)\(=\frac{11}{7}\)

\(\frac{6^5+6^6.7}{6^3+6^6.4}=\frac{6^5.\left(1+6.7\right)}{6^3.\left(1+6^3.4\right)}\)

\(=\frac{6^2.43}{865}=\frac{36.43}{865}\)

\(=\frac{1568}{865}\)