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\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x-3\right)\left(x+2\right)\)
\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)\(x^3-19x-30=\left(x^3+8\right)-\left(19x-38\right)=\left(x+2\right)\left(x^2-2x+4\right)-19\left(x+2\right)=\left(x+2\right)\left(x^2-2x-15\right)=\left(x+2\right)\left(x^2-5x+3x-15\right)=\left(x+2\right)\left(x-5\right)\left(x+3\right)\)
\(x^4+4x^2-5=x^4+4x^2+4-9=\left(x^2+2\right)^2-9=\left(x^2+5\right)\left(x^2-1\right)=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
\(x^3-7x-6=0\Leftrightarrow\left(x^3+1\right)-\left(7x+7\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\\x=-1\end{matrix}\right.\)
\(x^3-3x^2-16x+48=x^2\left(x-3\right)-16\left(x-3\right)=\left(x^2-16\right)\left(x-3\right)=\left(x-4\right)\left(x+4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\\x=-4\end{matrix}\right.\)
Bài 3:
a: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
=-5n chia hết cho 5
b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)
\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)
\(=n^2+3n-4-\left(n^2-3n-4\right)\)
\(=6n⋮6\)
Bài 1:
b:
x=9 nên x+1=10
\(M=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-x^7\left(x+1\right)+...-x\left(x+1\right)+x+1\)
\(=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^2-x+x+1\)
=1
c: \(N=\left(1+2+2^2+2^3+2^4\right)+2^5\left(1+2+2^2+2^3+2^4\right)+2^{10}\left(1+2+2^2+2^3+2^4\right)\)
\(=31\left(1+2^5+2^{10}\right)⋮31\)
Câu 2:
Tham khảo ở đây
Câu hỏi của Le Thi Hong Van - Toán lớp 6 - Học toán với OnlineMath