Trước hết ta có \(\left(\dfrac{x}{y}\right)^{-z}=\dfrac{1}{\left(\dfrac{x}{y}\right)^z}=\dfrac{1}{\dfrac{x^z}{y^z}}=\dfrac{y^z}{x^z}\)

Suy ra:

\(A=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)

\(A=\left(\dfrac{1}{4}\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)

\(A=4\cdot4^2\cdot\dfrac{3^2}{4^2}\cdot\dfrac{4}{5}\cdot\dfrac{3^3}{2^3}=4^2\cdot3^5\text{​​}\div5\div2^3\)

\(A=2^4\div2^3\cdot3^5\div5=2\cdot3^5\div5=2\cdot243\div5=\dfrac{486}{5}\)

Ta có\(\dfrac{5x}{7y}=\dfrac{-1}{3}\Leftrightarrow\dfrac{x}{y}=\dfrac{-7}{15}\Leftrightarrow\dfrac{x}{-7}=\dfrac{y}{15}\)

Áp dụng dãy tỉ số bằng nhau 

\(\dfrac{x}{-7}=\dfrac{y}{15}=\dfrac{-2x}{14}=\dfrac{3y}{45}=\dfrac{-2x+3y}{14+45}=\dfrac{7}{59}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{-7}=\dfrac{7}{59}\\\dfrac{y}{15}=\dfrac{7}{59}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{49}{59}\\y=\dfrac{105}{59}\end{matrix}\right.\)

\(\dfrac{x}{7}=\dfrac{5}{6}\Rightarrow x=\dfrac{7\cdot5}{6}=\dfrac{35}{6}\\ Vậy\text{ }x=\dfrac{35}{6}\)

\(\left(\dfrac{1}{2}\right)^{12}=\left(\dfrac{1}{8}\right)^3\\ \left(\dfrac{1}{3}\right)^9=\left(\dfrac{1}{27}\right)^3\\ Ta\text{ }có:\dfrac{1}{8}>\dfrac{1}{27}\\ Vậy:\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)

\(\left(\dfrac{1}{2}\right)^{12}=\dfrac{1}{4096};\left(\dfrac{1}{3}\right)^9=\dfrac{1}{19683}\\ \Rightarrow\dfrac{1}{4096}>\dfrac{1}{19683}\\ \Rightarrow\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)

cho : b² = ac

đem vào bên trái CMR , tính ra bên phải

\(Theo\text{ }bài\text{ }ra:2a=3b=4c\\ \Rightarrow\dfrac{2a}{12}=\dfrac{3b}{12}=\dfrac{4c}{12}\\ \Rightarrow\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}\\ \RightarrowĐặt\text{ }\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=k\\ \Rightarrow\left\{{}\begin{matrix}a=6k\\b=4k\\c=3k\end{matrix}\right.\\ Khi\text{ }đó\dfrac{a-b+c}{a+2b-c}=\dfrac{6k-4k+3k}{6k+8k-3k}=\dfrac{5k}{11}=\dfrac{5}{11}\\ Vậy:A=\dfrac{5}{11}.\)