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\(\frac{\left(-4\right)^6.9^5-\left(-6\right)^9.120.1^{2015}}{8^4.3^{12}-6^{11}.2016^0}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5.1}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}.1}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{4}{5}\)
\(\frac{\left(-4\right)^6.9^5-\left(-6\right)^9.120.1^{2015}}{8^4.3^{12}-6^{11}.2016^0}=\frac{\left(-2\right)^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{4}{5}\)
Ta có : $1-3-5-7-9-...-49$
$=1-(3+5+7+9+...+49)$
$=1-\dfrac{(3+49).24}{2}$
$=1-624=-623$
\(1-3-5-7-...-49\)
\(=1+\left(-3\right)+\left(-5\right)+\left(-7\right)+...+\left(-49\right)\)
Đặt:
\(A=\left(-3\right)+\left(-5\right)+\left(-7\right)+...+\left(-49\right)\)
\(A=\left[\left(\dfrac{49-3}{2}+1\right)\right]:2\left[\left(-3\right)+\left(-49\right)\right]\)
\(A=-624\)
Thay vào ta có giá trị biểu thức ban đầu là:
\(1+\left(-624\right)=-623\)
\(\frac{2^{4+6}}{2^{5.2}}-\frac{2^5.3^3.5^3}{2^3.3^3.2^2.5^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.3^3.5^3}{2^5.3^3.5^2}=1-5=-4\)
Bài 2:
\(B=\dfrac{2^{15}\cdot5^8-2^5\cdot2^9\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{5}{4}\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)
a)\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)\)
\(=\frac{5}{9}:\left[\left(\frac{1}{11}-\frac{5}{22}\right)+\left(\frac{1}{15}-\frac{2}{3}\right)\right]\)
\(=\frac{5}{9}:\left[-\frac{3}{22}+-\frac{3}{5}\right]\)
\(=\frac{5}{9}:\frac{-81}{110}\)
\(=\frac{-550}{729}\)