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\(\frac{5^3.2.3^2.5.2^6}{5^{10}.3^2.2^{13}}=\frac{5^4.2^7.3^2}{5^{10}.3^2.2^{13}}=\frac{1}{5^6.2^6}=\frac{1}{10^6}\)
\(\frac{18\left(27-23\right)}{4\left(34-52\right)}=\frac{18.4}{4.\left(-18\right)}=-1\)
\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
\(\frac{2x+1}{3}=\frac{5}{2}\)
\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)
2x= 15/2 - 1 = 13/2
x = 13/2 : 2
x = 13/4
b) 2x + 2x+1 + 2x+2 + 2x+3 = 480
2x.(1+ 2 +22 + 23) = 480
2x . 15 = 480
2x = 480 : 15 = 32
2x = 25 => x = 5
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)
< = > 3x= -6 => x = -2
\(\left(\frac{1}{1}+\frac{1}{2}\right).\left(\frac{1}{1}+\frac{1}{3}\right).\left(\frac{1}{1}+\frac{1}{4}\right)...\left(\frac{1}{1}+\frac{1}{2016}\right).\left(\frac{1}{1}+\frac{1}{2017}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2017}{2016}.\frac{2018}{2017}\)
\(=\frac{3.4.5...2017.2018}{2.3.4...2016.2017}\)
\(=\frac{2018}{2}=1009\)
b) Ta có:
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(\Rightarrow B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(\Rightarrow B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(\Rightarrow B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Vậy \(\frac{A}{B}=\frac{1}{2017}\)
Gọi d=ƯCLN(6n+12;3n+5).
Ta có:6n+12 chia hết cho d. suy ra: 3n+6 chia hết cho d.
3n+5 chia hết cho d.
suy ra: (3n+6)-(3n+5) chia hết cho d.
suy ra: 1 chia hết cho d.
suy ra: d=1.
Vậy \(\frac{6n+12}{3n+5}\) là PS tối giản.
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)
Xét \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
và \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.10=\frac{10}{19}>\frac{1}{2}\)
Do đó \(B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=\frac{5}{4}>1\)