1, (x2-x+2)2-(x-2)2=(x2-x+2-x+2)(x2-x+2+x-2)=(x2-2x+4)x2
2,a.x3+4x2-29x+24=0
\(\Leftrightarrow\)x3-3x2+7x2-21x-8x+24=0
\(\Leftrightarrow\)(x3-3x2)+(7x2-21x)-(8x+24)=0
\(\Leftrightarrow\)x2(x-3)+7x(x-3)-8(x-3)=0
\(\Leftrightarrow\)(x-3)(x2-x+8x-8)=0
\(\Leftrightarrow\)(x-3)(x-1)(x+8)=0
\(\Leftrightarrow\)\(\left[\begin{matrix}x-3=0\\x-1=0\\x+8=0\end{matrix}\right.\)\(\left[\begin{matrix}x=3\\x=1\\x=-8\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-8;1;3\right\}\)
b. đặt x2-x=y ta có:
y2-14y+24=0 \(\Leftrightarrow\)(y2-2.7y+49)-25=0 \(\Leftrightarrow\)(y-7)2-52=0 \(\Leftrightarrow\)(y-12)(y-2)=0 \(\Leftrightarrow\left[\begin{matrix}y=12\\y=2\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[\begin{matrix}x^2-x=12\\x^2-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x^2-x-12=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\left(x+3\right)\left(x-4\right)=0\\\left(x-2\right)\left(x+1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-3\\x=4\\x=2\\x=-1\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-3;-1;2;4\right\}\)
