\(1.x^4+6x^3+11x^2+6x+1\)
\(=x^4+6x^3+9x^2+2x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+2x^2+6x\)
\(=\left(x^2\right)^2+\left(3x\right)^2+1^2+2.x^2.3x+2.x^2.1+2.3x.1\)
\(=\left(x^2+3x+1\right)^2\)
\(2,6x^4+5x^3-38x^2+5x+6\)
\(=6x^4+6x^3+2x^3-3x^3-36x^2+2x^2-3x^2-x^2-12x+18x-x+6\)
\(=\left(6x^4+2x^3\right)+\left(6x^3+2x^2\right)-\left(3x^3+x^2\right)-\left(36x^2+12x\right)+\left(18x+6\right)-\left(3x^2+x\right)\)
\(=2x^3\left(3x+1\right)+2x^2\left(3x+1\right)-x^2\left(3x+1\right)-12x\left(3x+1\right)+6\left(3x+1\right)-x\left(3x+1\right)\)
\(=\left(3x+1\right)\left(2x^3+2x^2-x^2-12x+6-x\right)\)
\(=\left(3x+1\right)\left[\left(2x^3-x^2\right)+\left(2x^2-x\right)-\left(12x-6\right)\right]\)
\(=\left(3x+1\right)\left[x^2\left(2x-1\right)+x\left(2x-1\right)-6\left(2x-1\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+3x-2x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left[\left(x^2+3x\right)-\left(2x+6\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x+3\right)\left(x-2\right)\)
1. \(x^4+6x^3+11x^2+6x+1\)
\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2+2x^2+6x+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
3. \(x^4-7x^3+14x^2-7x+1\)
\(=x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)
\(=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+14\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right).\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{1}{4}\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}-\dfrac{7}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=\left(x^2+1-\dfrac{7}{2}x\right)^2-\left(\dfrac{1}{2}x\right)^2\)
\(=\left(x^2-3x+1\right)\left(x^2-4x+1\right)\)
Có thể phân tích thành HĐT tiếp hoặc không.
\(1.\text{ }x^4+6x^3+11x^2+6x+1\)
Dễ thấy đa thức trên sau khi phân tích thành nhân tử sẽ có dạng:
\(\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ =x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\\ =x^4+\left(c+a\right)x^3+\left(d+ac+b\right)x^2+\left(ad+bc\right)x+bd\)
Đồng nhất da thức tren với đa thức đã cho
\(\text{Ta được: }\left\{{}\begin{matrix}c+a=6\\d+ac+b=11\\ad+bc=6\\bd=1\Rightarrow b=1;d=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}c+a=6\\ac=9\Rightarrow a=3;c=3\\a+c=6\end{matrix}\right.\)
Từ \(a=3;b=1;c=3;d=1\) suy ra:
\(x^4+6x^3+11x^2+6x+1\\ =\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ =\left(x^2+3x+1\right)\left(x^2+3x+1\right)\\ =\left(x^2+3x+1\right)^2\)\(2.\text{ }6x^4+5x^3-38x^2+5x+6\)
Dễ thấy đa thức trên sau khi phân tích thành nhân tử sẽ có dạng: \(\left(ax^2+bx+c\right)\left(dx^2+ex+f\right)\\ =adx^4+aex^3+afx^2+bdx^3+bex^2+bfx+cdx^2+cex+cf\\ =adx^4+\left(ae+bd\right)x^3+\left(af+be+cd\right)x^2+\left(bf+ce\right)x+cf\)
Đồng nhất da thức tren với đa thức đã cho
\(\text{Ta được: }\left\{{}\begin{matrix}ad=6\Rightarrow a=2;d=3\\ae+bd=5\\af+be+cd=-38\\bf+ce=5\\cf=6\Rightarrow c=2;f=3\end{matrix}\right.\\ \left\{{}\begin{matrix}2e+3b=5\\be=-50\Rightarrow e=-10;b=5\\3b+2e=5\end{matrix}\right.\)
Từ \(a=2;b=5;c=2;d=3;e=-10;f=3\) suy ra :
\(6x^4+5x^3-38x^2+5x+6\\ =\left(ax^2+bx+c\right)\left(dx^2+ex+f\right)\\ =\left(2x^2+5x+2\right)\left(3x^2-10x+3\right)\\ =\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)\\ =\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]\\ =\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]\\ =\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)\)
\(3.\text{ }x^4-7x^3+14x-7x+1\)
Dễ thấy đa thức trên sau khi phân tích thành nhân tử sẽ có dạng: \(\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ =x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\\ =x^4+\left(c+a\right)x^3+\left(d+ac+b\right)x^2+\left(ad+bc\right)x+bd\)
Đồng nhất da thức tren với đa thức đã cho
\(\text{Ta được: }\left\{{}\begin{matrix}c+a=-7\\d+ac+b=14\\ad+bc=-7\\bd=1\Rightarrow b=1;d=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}c+a-7\\ac=12\Rightarrow a=-4;c=-3\\a+c=-7\end{matrix}\right.\)
Từ \(a=-4;b=1;c=-3;d=1\) suy ra :
\(x^4-7x^3+14x^2-7x+1\\ =\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ =\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)
nhờ mn là giúp mình với ạ , minh đang cần gấp :( 
