a.
Xét pt:
\(x^3-y=x\left(xy-1\right)\)
\(\Leftrightarrow x^3+x-x^2y-y=0\)
\(\Leftrightarrow x\left(x^2+1\right)-y\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x=y\)
Thế vào pt dưới:
\(x^4-7x^2+4x+5=0\)
\(\Leftrightarrow\left(x^2-x-1\right)\left(x^2+x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-5=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
Xét pt: \(2x^3+x=2x^2y+y\)
\(\Leftrightarrow x\left(2x^2+1\right)=y\left(2x^2+1\right)\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow x=y\)
Thay vào pt dưới:
\(\left(4x+3\right)^2\left(2x+1\right)\left(x+1\right)=810\)
\(\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+3x+1\right)=810\)
Đặt \(2x^2+3x+1=t\Rightarrow16x^2+24x+9=8t+1\)
Pt trở thành:
\(\left(8t+1\right)t=810\Leftrightarrow8t^2+t-810=0\)
\(\Rightarrow\left[{}\begin{matrix}t=10\\t=-\dfrac{81}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x^2+3x+1=10\\2x^2+3x+1=-\dfrac{81}{8}\end{matrix}\right.\)
\(\Leftrightarrow...\)
hihi, chủ đề là gì mình bấm bừa nha