-3/7-1/4.3/7+3/7.5/4
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\(a) 5^{n+1}+7.5^n+5.7^{n+2}+7^{n+3}\\ =5^n . 5+7.5^n+5.7^{n+2}+7^{n+2}.7\\ =5^n( 5+7)+7^{n+2}(5+7)\\ =5^n.12+7^{n+2}.12\\ =12.(5^n+7^{n+2})\)
Vì 12 ⋮ 2
=> 12.5n + 7n+2 ⋮ 2
Vậy \( 5^{n+1}+7.5^n+5.7^{n+2}+7^{n+3}\\\)⋮ 2
\(b) 3^{n+1}+4^{b+1}+3.4^b+4.3^n\\ =3^n.3+4^b.4+3.4^b+4.3^n\\ =(4^b.4+3.4^b)+(3^n.3+4.3^n)\\ =4^b(4+3)+3^n(3+4)\\ =4^n.7+3^n.7\\ =7.(4^n+3^n)\)
Vì 7 ⋮ 7
=>7.(4n + 3n) ⋮ 7
Vậy \(3^{n+1}+4^{b+1}+3.4^b+4.3^n\\\)⋮ 7
c: \(C=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\dfrac{9^3}{4^3}:\dfrac{3^3}{16^3}}{2^7\cdot5^2+2^9}=\dfrac{1+1728}{3712}=\dfrac{1729}{3712}\)
\(D=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\dfrac{3^5-3^4}{3^6+3^5}=\dfrac{3^4\left(3-1\right)}{3^5\left(3+1\right)}=\dfrac{2}{3\cdot4}=\dfrac{2}{12}=\dfrac{1}{6}\)
\(E=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}=\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}=5\cdot\dfrac{-2}{3}=\dfrac{-10}{3}\)
\(P=\dfrac{16^7\cdot5^3\left(5-1\right)}{16^7\cdot\left(25^2-5^3\right)}=\dfrac{5^3\cdot2^2}{5^4-5^3}=\dfrac{5^3\cdot2^2}{5^3\cdot\left(5-1\right)}=1\)
\(F=\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^9\div\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
\(F=\frac{\left(\frac{2.5}{5}\right)^7+\left(\frac{9.16}{4.3}\right)^3}{2^7.5^2+2^9}=\frac{2^7+12^3}{2^7.5^2+2^9}=\frac{2^7+2^6.3^3}{2^7.5^2+2^9}=\frac{2^6.\left(2+3^3\right)}{2^7.\left(5^2+2^2\right)}=\frac{2^6.29}{2^7.29}\)
\(F=\frac{1}{2}\)
\(-\dfrac{3}{7}-\dfrac{1}{4}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{5}{4}\\ =\dfrac{3}{7}.\left(-1-\dfrac{1}{4}+\dfrac{5}{4}\right)\\ =\dfrac{3}{7}.0\\ =0\)
\(\dfrac{-3}{7}-\dfrac{1}{4}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{5}{4}\)
=\(\dfrac{-3}{7}-\dfrac{3}{28}+\dfrac{15}{28}\)
=\(\dfrac{-15}{28}+\dfrac{15}{28}\)
=\(0\)
\(#LilyVo\)