help me

dễ mà bạn xin 20 phút làm ra giấy nhé :)) 

a) \(\left(x^4+6x^3+9x^2\right)+2x^2+6x+1\)

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

        \(\left(x^2+3x+1\right)^2\)

b) \(x^4+x^3+x^2+x+1\)

câu b,  chúa sẽ c/m x ko tồn tại , và nó là 1 đa thức bất khả Q . trong R 

vì lớp 8 chưa học đến số phức  

     \(x^4+x^3=-x^2-x-1\)

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

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

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

\(\Delta`=\left(-2\right)^2-\left(-4\right).\left(-3\right)=4-12< 0\)

          denta < 0 x vô nghiệm

vậy đa thức trên ko thể phân tích và nó là 1 đa thức bất khả Q

c) ,   

\(\left(6x^4-12x^3\right)+\left(17x^3-34x^2\right)-\left(4x^2-8x\right)-\left(3x-6\right)\)

\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)

     \(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)\)

      \(\left(x-2\right)\left\{\left(6x^3+18x^2\right)-\left(x^2+3x\right)-\left(x+3\right)\right\}\)

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

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

 \(\left(x-2\right)\left(x+3\right)\left\{\left(6x^2+\frac{6}{3}x\right)-\left(\frac{9}{3}x+\frac{9}{9}\right)\right\}\)

\(\left(x-2\right)\left(x+3\right)\left\{6x\left(x+\frac{1}{3}\right)-\frac{9}{3}\left(x+\frac{1}{3}\right)\right\}\)

\(\left(X-2\right)\left(X+3\right)\left(X+\frac{1}{3}\right)\left(6x-1\right)\)

d)

\(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)\)

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

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

\(\left(x-1\right)\left\{\left(x^3-x^2\right)+\left(7x^2-7x\right)+\left(x-1\right)\right\}\)

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

\(\Delta=49-4=45\)

\(x1,2=\frac{-7+\sqrt{45}}{2},\frac{-7-\sqrt{45}}{2}\)

\(\left(x-1\right)^2\left(x-\frac{7+\sqrt{45}}{2}\right)\left(x-\frac{7-\sqrt{45}}{2}\right)\)

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 :

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

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

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)

a, \(x^4-6x^3+11x^2-6x+1=0\)

\(\Rightarrow\left(x^2-3x+1\right)^2=0\)

\(\Rightarrow x^2-3x+1=0\)

\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)

Chúc bạn học tốt

\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)

\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)

\(\left(x^2-3x+1\right)^2=0\)

tự làm

B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)

\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)

\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)

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

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

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

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

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

\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)

câu C nghĩ đã

8X-5(5X^2-1)^2=-4

a, \(x^4-6x^3+11x^2-6x+1=0\)

=> \(x^4-6x^3+9x^2+2x^2-6x+1=0\)

=> \(x^2+3x+1=0\)

=> \(\Delta\) =\(b^2-4c\)

=\(3^2.4=5\)

Nên \(\sqrt{\Delta}=5\)

x= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+\sqrt{5}}{2}\)

hoặc x= \(\dfrac{b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{5}}{2}\)

Đáp án câu a.

https://giaibaitapvenha.blogspot.com/2017/12/toan-lop-8-ai-so_27.html

       \(4x^4+4x^3+5x^2+2x+1\)

\(=\left(2x^2\right)^2+2.2x^2.x+x^2+4x^2+2x+1\)

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

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

       \(x^4+6x^3+11x^2+6x+1\)

\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2+2x^2+6x+1\)

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

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

Chúc bạn học tốt.