Trường hợp 1: x=1,5 và y=-0,75

\(P=\dfrac{3}{2}-4\cdot\dfrac{3}{2}\cdot\dfrac{-3}{4}+\dfrac{-3}{4}=\dfrac{3}{4}+3\cdot\dfrac{3}{2}=\dfrac{3}{4}+\dfrac{9}{2}=\dfrac{3}{4}+\dfrac{18}{4}=\dfrac{21}{4}\)

Trường hợp 2: x=-1,5 và y=-0,75

\(P=\dfrac{-3}{2}-4\cdot\dfrac{-3}{2}\cdot\dfrac{-3}{4}+\dfrac{-3}{4}\)

\(=\dfrac{-9}{4}-4\cdot\dfrac{9}{8}=\dfrac{-9}{4}-\dfrac{9}{2}=\dfrac{-27}{4}\)

Lời giải:

\(|x|=1,5\Rightarrow \left[\begin{matrix} x=1,5\\ x=-1,5\end{matrix}\right.\)

Nếu $x=1,5$ và $y=-0,75$:

\(P=x-4xy+y=1,5-4.1,5.(-0,75)+(-0,75)=\frac{21}{4}\)

Nếu \(x=-1,5; y=-0,75\):

\(P=x-4xy+y=(-1,5)-4(-1,5)(-0,75)+(-0,75)=\frac{-27}{4}\)

Vì \(\left|x\right|=1,5\)

\(\Rightarrow x=1,5\) hoặc \(x=-1,5\)

Thay \(x=1,5;y=-0,75\) vào P:

\(P=1,5-4.1,5.\left(-0,75\right)+\left(-0,75\right)\)

\(=5,25\)

Thay \(x=-1,5;y=-0,75\) vào P ta đc:

\(P=-1,5-4.\left(-1,5\right).\left(-0,75\right)+\left(-0,75\right)\)

\(=-6,75\)

Vậy P \(\left[\begin{matrix}=5,25\\=-6,75\end{matrix}\right.\)

\(x_1=2;y_1=3\\ \Rightarrow y_1=\dfrac{3}{2}x_1\\ \Rightarrow y_2=\dfrac{3}{2}x_2\)

Mà \(3x_2+5y_2=10\)

\(\Rightarrow3x_2+\dfrac{3}{2}\cdot5x_2=10\\ \Rightarrow x_2\left(3+\dfrac{15}{2}\right)=10\\ \Rightarrow x_2=10:\dfrac{21}{2}=\dfrac{20}{21}\\ \Rightarrow y_2=\dfrac{20}{21}\cdot\dfrac{3}{2}=\dfrac{10}{7}\)

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

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

\(=\frac{324}{756}\)

=\(\frac{3}{7}\)