Không biết em có làm sai không:
ĐKXĐ: \(x,y\ge0\).
Đặt 2x = a; 3y = b.
HPT trở thành:
\(\left\{{}\begin{matrix}\left(\sqrt{5}\right)^a-\left(\sqrt{5}\right)^b+\left(a-b\right)\left(ab+12\right)=0\\a^2+b^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=16\\\left(\sqrt{5}\right)^a-\left(\sqrt{5}\right)^b+\left(b-a\right)\left(a^2+b^2\right)+a^3-b^3+12\left(a-b\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=16\\\left(\sqrt{5}\right)^a+a^3-4a=\left(\sqrt{5}\right)^b+b^3-4b=0\left(1\right)\end{matrix}\right.\).
Giả sử \(a\ge b\Rightarrow\left(\sqrt{5}\right)^a\ge\left(\sqrt{5}\right)^b\). Mà \(\left(a^3-4a\right)-\left(b^3-4b\right)=\left(a-b\right)\left(a^2+ab+b^2-4\right)\ge0\) nên VT(1) \(\ge\) VP(1).
Do đẳng thức xảy ra nên ta có a = b. Thay vào ta tìm được a = b = \(2\sqrt{2}\) nên \(x=\sqrt{2};y=\dfrac{2\sqrt{2}}{3}\).
\(\left\{{}\begin{matrix}\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=\left(3y-2x\right)\left(6xy+12\right)\left(1\right)\\4x^2+9y^2=16\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Rightarrow4x^2+9y^2-4=12\) the vo (1)
\(\Rightarrow\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=\left(3y-2x\right)\left(6xy+4x^2+9y^2-4\right)\)
\(\Leftrightarrow\left(\sqrt{5}\right)^{2x}-\left(\sqrt{5}\right)^{3y}=27y^3-8x^3-12y+8x\)
\(\Leftrightarrow\left(\sqrt{5}\right)^{2x}+\left(2x\right)^3-4.\left(2x\right)=\left(\sqrt{5}\right)^{3y}+\left(3y\right)^3-4.\left(3y\right)\left(3\right)\)
Xét hàm số \(f\left(t\right)=\left(\sqrt{5}\right)^{2t}+\left(2t\right)^3-4.2t\) đồng biến trên R
\(\Rightarrow\left(3\right):f\left(2x\right)=f\left(3y\right)\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\4x^2+9y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=\dfrac{2\sqrt{2}}{3}\end{matrix}\right.\)
