Câu 1 : Tìm x , biết
a) x3-3x2-16x+48=0
b) x3-7x-6=0
c) 2x4-x3+2x3+3x-2=0
Câu 2 : Phân tích thành nhân tử
a) A=x2-x-6
b) B=x4+4x2-5
c) C=x3-19x-30
d) D = x4+x2+1
Câu 13: Chứng minh rằng
a) 2n3-3n2+n chia hết cho 6 , vs mọi giá trị nguyên n
b) n3+11n chia hết chó 6 , vs n nguyên
c) n4+2n3+3n2+2n chia hết cho 8 , vs n nguyên
d) 2n3+3n2+7n chia hết cho 6 , với n nguyên
\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)\)
\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)\(x^3-19x-30=\left(x^3+8\right)-\left(19x-38\right)=\left(x+2\right)\left(x^2-2x+4\right)-19\left(x+2\right)=\left(x+2\right)\left(x^2-2x-15\right)=\left(x+2\right)\left(x^2-5x+3x-15\right)=\left(x+2\right)\left(x-5\right)\left(x+3\right)\)
\(x^4+4x^2-5=x^4+4x^2+4-9=\left(x^2+2\right)^2-9=\left(x^2+5\right)\left(x^2-1\right)=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(x^3-7x-6=0\Leftrightarrow\left(x^3+1\right)-\left(7x+7\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=-1\end{matrix}\right.\)
\(x^3-3x^2-16x+48=x^2\left(x-3\right)-16\left(x-3\right)=\left(x^2-16\right)\left(x-3\right)=\left(x-4\right)\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-4\end{matrix}\right.\)
\(2x^4-x^3+2x^2+3x-2=0\Leftrightarrow2x^4-x^3+2x\left(x+1\right)-2\left(x+1\right)+3x=0\Leftrightarrow3x^4+3x-x^3\left(x+1\right)+2x\left(x+1\right)-2\left(x+1\right)=0\Leftrightarrow3x\left(x+1\right)\left(x^2-x+1\right)-x^3\left(x+1\right)+2x\left(x+1\right)-2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x^3-3x^2+5x-2\right)=0\)