B1:
Ta có: \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}-2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.\left(-4\right)}{3.5}=-\frac{8}{15}\)
B2:
Ta có: \(1+3+5+...+x=1600\)
\(\Leftrightarrow\frac{\left(x+1\right)\cdot\left(\frac{x-1}{2}+1\right)}{2}=1600\)
\(\Leftrightarrow\left(x+1\right)\cdot\frac{x+1}{2}=3200\)
\(\Leftrightarrow\left(x+1\right)^2=6400\)
Xét theo dãy tăng tiến ta thấy được giá trị của x càng tăng
=> x dương => x + 1 dương
\(\Rightarrow x+1=80\)
\(\Rightarrow x=79\)
B3:
Ta có: \(15^{12}=\left(3.5\right)^{12}=3^{12}.5^{12}\)
và \(81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}\)
Mà \(3^{12}.5^{15}>3^{12}.5^{12}\)
=> \(15^{12}< 81^3.125^5\)
