Question

 bài 1 
\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
2. Tìm ba số x, y, z biết rằng 
\(\frac{x}{2}=\frac{y}{3},\frac{y}{4}=\frac{z}{5}\) và\(x+y+z=10\)
Nhờ các bạn giải giúp mk !

Answer 1

\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)

\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)

\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)

P/s: Hoq chắc ạ (: Ms lp 6 lm đại

Answer 2

\(\frac{x}{2}=\frac{y}{3}\)

\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)

\(\frac{y}{4}=\frac{z}{5}\)

\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)

Từ (1) (2)

 \(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)