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\(8x^3-1=\left(2x\right)^3-1^3=\left(2x-1\right)\left[\left(2x\right)^2+1.2x+1^2\right]\)
a: \(9x^2-6x+3\)
\(=\left(9x^2-6x+1\right)+2\)
\(=\left(3x-1\right)^2+2\ge2\)
b: \(6x-x^2+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left(x-3\right)^2+10\le10\)
\(x^3+9x^2+11x-21=0\)
\(\Leftrightarrow x^3+7x^2+2x^2+11x-21=0\)
\(\Leftrightarrow x^2\left(x+7\right)+2x^2+11x-21=0\)
\(\Leftrightarrow x^2\left(x+7\right)+x^2+7x+x^2+4x-21=0\)
\(\Leftrightarrow x^2\left(x+7\right)+x\left(x+7\right)+x^2+4x-21=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x^2+4x-21=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x^2+7x-3x-21=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x\left(7+x\right)-3\left(7+x\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-3\\x=1\end{matrix}\right.\)
cách khác: "định hướng HĐT"
\(\left(x^3+3.3.x^2+3.3^2x+3^3\right)+\left[\left(-27x+11x\right)-27-21\right]=\left(x+3\right)^3-16\left(x+3\right)=0\)\(\left(x+3\right)\left[\left(x+3\right)^2-16\right]=\left(x+3\right)\left[\left(x+3\right)-4\right]\left[\left(x+3\right)+4\right]\)
\(\left(x+3\right)\left(x-1\right)\left(x+7\right)=0\)
\(\left[{}\begin{matrix}x=-3\\x=1\\x=-7\end{matrix}\right.\)
\(A=\left(2n-1\right)^3-2n+1\)
\(A=8n^3-6n+6n-1-2n+1\)
\(A=8n^3-2n=2n\left(4n^2-1\right)\)
\(A=2n\left(2n+1\right)\left(2n-1\right)\)
\(A=\left(2n-1\right)2n\left(2n+1\right)⋮6\) ( 3 số tự nhiên liên tiếp)