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Sửa đề: \(C=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\right)^6\cdot3^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
\(C=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\)
\(=\dfrac{2}{3\cdot4}-\dfrac{5\cdot\left(-6\right)}{9}\)
\(=\dfrac{2}{12}+\dfrac{30}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{1}{6}+\dfrac{20}{6}=\dfrac{21}{6}=\dfrac{7}{2}\)
\(a)A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\dfrac{2^{12}.3^5-\left(2^2\right)^63.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(A=\dfrac{2^{12}.3^5-2^{12}.3^5}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.7^3.2^3}\)
\(A=\dfrac{0}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+5^3+2^3\right)}\)
\(A=0-\dfrac{5^4.\left(-6\right)}{1+125+8}\)
\(A=0-\dfrac{625.\left(-6\right)}{134}\)
\(A=\dfrac{-3750}{134}\)\(=\dfrac{-1875}{67}\)
\(b)3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.3^2-2^n.2^2+3^n-2^n\)
\(=(3^n.9+3^n)-\left(2^n.4+2^n\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\)
\(Suy\) \(ra:\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
b. Ta có: \(3^{n +2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=\left(3^n.3^2+3^n\right)-\left(2^{n-1}.2^3+2^{n-1}.2\right)\)
\(=3^n.\left(3^2+1\right)-2^{n-1}\left(2^3+2\right)\)
\(=3^n.10-2^{n-1}.10⋮10\)
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(B=\frac{2^{12}.3^6-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(B=\frac{2^{12}.3^4.\left(3^2-1\right)}{2^{12}.3^5.\left(3-1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(B=\frac{8}{6}-\frac{5-35}{9}\)
\(B=\frac{4}{3}+\frac{30}{9}\)
\(B=\frac{4}{3}+\frac{10}{3}\)
\(B=\frac{14}{3}\)
vậy \(B=\frac{14}{3}\)