help me

a)\(x^4+6x^3+11x^2+6x+1\)

\(=x^4+9x^2+1+6x^3+6x+2x^2\)

\(=\left(x^2+3x+1\right)^2\)

\(x^4+5x^3-12x^2+5x+1\)

\(=\left(x^4-2x^3+x^2\right)+\left(7x^3-14x^2+7x\right)+\left(x^2-2x+1\right)\)

\(=x^2\left(x^2-2x+1\right)+7x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)\)

\(=\left(x^2+7x+1\right)\left(x^2-2x+1\right)\)

\(=\left(x^2+7x+1\right)\left(x-1\right)^2\)

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

\(1.x^4+6x^3+11x^2+6x+1\)

\(=x^4+6x^3+9x^2+2x^2+6x+1\)

\(=x^4+9x^2+1+6x^3+2x^2+6x\)

\(=\left(x^2\right)^2+\left(3x\right)^2+1^2+2.x^2.3x+2.x^2.1+2.3x.1\)

\(=\left(x^2+3x+1\right)^2\)

\(2,6x^4+5x^3-38x^2+5x+6\)

\(=6x^4+6x^3+2x^3-3x^3-36x^2+2x^2-3x^2-x^2-12x+18x-x+6\)

\(=\left(6x^4+2x^3\right)+\left(6x^3+2x^2\right)-\left(3x^3+x^2\right)-\left(36x^2+12x\right)+\left(18x+6\right)-\left(3x^2+x\right)\)

\(=2x^3\left(3x+1\right)+2x^2\left(3x+1\right)-x^2\left(3x+1\right)-12x\left(3x+1\right)+6\left(3x+1\right)-x\left(3x+1\right)\)

\(=\left(3x+1\right)\left(2x^3+2x^2-x^2-12x+6-x\right)\)

\(=\left(3x+1\right)\left[\left(2x^3-x^2\right)+\left(2x^2-x\right)-\left(12x-6\right)\right]\)

\(=\left(3x+1\right)\left[x^2\left(2x-1\right)+x\left(2x-1\right)-6\left(2x-1\right)\right]\)

\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+x-6\right)\)

\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+3x-2x-6\right)\)

\(=\left(3x+1\right)\left(2x-1\right)\left[\left(x^2+3x\right)-\left(2x+6\right)\right]\)

\(=\left(3x+1\right)\left(2x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)

\(=\left(3x+1\right)\left(2x-1\right)\left(x+3\right)\left(x-2\right)\)

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 :

\(x^4-7x^3+14x^2-7x+1\\ =\left(x^2+ax+b\right)\left(x^2+cx+d\right)\\ =\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)

2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

=(2x+5)(x-1)

8: =(x^2-1)*(x^2-4)

=(x-1)(x+1)(x-2)(x+2)

9: =x^2(x-5)-9(x-5)

=(x-5)(x-3)(x+3)

b) http://olm.vn/hoi-dap/question/118763.html

\(f,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Đặt \(t=x^2+5x+4\) , ta có
\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t^2+2t+1\right)-25\)

\(=\left(t+1\right)^2-5^2\)

\(=\left(t+1-5\right)\left(t+1+5\right)\)

\(=\left(t-4\right)\left(t+6\right)\)

\(=\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)

\(g,\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)

\(=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-20\)

\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)

Đặt \(t=x^2-8x+7\), ta có:

\(t\left(t+8\right)-20\)

\(=t^2+8t-20\)

\(=\left(t^2+8t+16\right)-36\)

\(=\left(t+4\right)^2-6^2\)

\(=\left(t+4+6\right)\left(t+4-6\right)\)

\(=\left(t+10\right)\left(t-2\right)\)

\(=\left(x^2-8x+7+10\right)\left(x^2-8x+7-2\right)\)

\(=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)

2:

a: \(x^2-12x+20\)

\(=x^2-2x-10x+20\)

=x(x-2)-10(x-2)

=(x-2)(x-10)

b: \(2x^2-x-15\)

=2x^2-6x+5x-15

=2x(x-3)+5(x-3)

=(x-3)(2x+5)

c: \(x^3-x^2+x-1\)

=x^2(x-1)+(x-1)

=(x-1)(x^2+1)

d: \(2x^3-5x-6\)

\(=2x^3-4x^2+4x^2-8x+3x-6\)

\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)

\(=\left(x-2\right)\left(2x^2+4x+3\right)\)

e: \(4y^4+1\)

\(=4y^4+4y^2+1-4y^2\)

\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)

\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)

f; \(x^7+x^5+x^3\)

\(=x^3\left(x^4+x^2+1\right)\)

\(=x^3\left(x^4+2x^2+1-x^2\right)\)

\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)

\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)

g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)

\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)

\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)

h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)

\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)

\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)

\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)

\(=\left(x+1\right)^4-4\left(x+1\right)^2\)

\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)

\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)

\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)

i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)

\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)

\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)

\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)

\(=\left(x+2y-1\right)\left(x+2y-3\right)\)

j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)

\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)

\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)

\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)