Lời giải:
a)
Đặt $x^2=a$ $(a\geq 0$) thì pt trở về dạng bậc 2: \(2a^2-a-1=0\)
\(\Leftrightarrow 2a(a-1)+(a-1)=0\Leftrightarrow (2a+1)(a-1)=0\)
\(\Rightarrow \left[\begin{matrix} a=-\frac{1}{2}< 0(\text{vô lý})\\ a=1(\text{chọn})\end{matrix}\right.\)
Với $a=1$ thì \(x^2=1\Leftrightarrow x=\pm 1\)
b)
Đặt \(x^2=a(a>0\) do $x\neq 0$)
PT \(\Leftrightarrow \frac{15}{a}-\frac{12}{a^2}=3\)
\(\Leftrightarrow 3a^2=15a-12\)
\(\Leftrightarrow 3a^2-15a+12=0\)
\(\Leftrightarrow 3(a-1)(a-4)=0\Rightarrow \left[\begin{matrix} a=1\\ a=4\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x^2=1\\ x^2=4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\pm 1\\ x=\pm 2\end{matrix}\right.\) (đều thỏa mãn)
Vậy..........
c)
\( \left\{\begin{matrix} x+y=3\\ y^2-x^2=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+y=3\\ (y-x)(y+x)=3\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y=3\\ y-x=1\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2y=(x+y)+(y-x)=4\\ 2x=(x+y)-(y-x)=2\end{matrix}\right. \Rightarrow \left\{\begin{matrix} y=2\\ x=1\end{matrix}\right.\)
Thay $y=2, x=1$ ta có:
\((y-x)^{2019}=(2-1)^{2019}=1\)
