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\(\frac{\left(-4\right)^6.9^5-\left(-6\right)^9.120.1^{2015}}{8^4.3^{12}-6^{11}.2016^0}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5.1}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}.1}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{4}{5}\)
\(\frac{\left(-4\right)^6.9^5-\left(-6\right)^9.120.1^{2015}}{8^4.3^{12}-6^{11}.2016^0}=\frac{\left(-2\right)^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{4}{5}\)
Bài 2:
\(B=\dfrac{2^{15}\cdot5^8-2^5\cdot2^9\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{5}{4}\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)
Ta có:
\(\left(x-1\right).\left(x+1\right).\left(x+2\right)\)
\(=\left(x^2+x-x-1\right).\left(x+2\right)\)
\(=\left(x^2-1\right).\left(x+2\right)\)
\(=x^3+2x^2-x-2\)
\(\left(-3,75\right)\left(-7,2\right)+2,8.3,75=3,75.7,2+2,8.3,75=\dfrac{15}{4}.\dfrac{36}{5}+\dfrac{14}{5}.\dfrac{15}{4}=\dfrac{15}{4}\left(\dfrac{36}{5}+\dfrac{14}{5}\right)=\dfrac{15}{4}.10=\dfrac{75}{2}=37.5\)