b) Tính
\(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)
Vậy : \(A=\frac{4}{7}\)
c)
Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{3.3}=\frac{y}{4.3}\Rightarrow\frac{x}{9}=\frac{y}{12}\) (1)
\(\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{3.4}=\frac{z}{5.4}\Rightarrow\frac{y}{12}=\frac{z}{20}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)
\(\Rightarrow\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\frac{2x}{18}=\frac{3y}{36}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{9}=3\\\frac{y}{12}=3\\\frac{z}{20}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=9.3\\y=12.3\\z=20.3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
Vậy : \(\left(x,y,z\right)=\left(27,36,60\right)\)
