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18 tháng 6 2019

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

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

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

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

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

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

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

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

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

8 tháng 9 2019

Câu a):

ta có (x2-x-2)2+(x-2)2

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

=(x-2)2(x2+2x+2)

Y
17 tháng 6 2019

a) \(A=\left(x^2+x-2\right)\left(x+7\right)-16\)

\(=x^3+8x^2+5x-14-16\)

\(=x^3+8x^2+5x-30\)

\(=x^3+3x^2+5x^2+15x-10x-30\)

\(=x^2\left(x+3\right)+5x\left(x+3\right)-10\left(x+3\right)\)

\(=\left(x^2+5x-10\right)\left(x+3\right)\)

b) \(A=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)

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

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

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

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

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

c) \(81x^4+4=81x^4+36x^2+4-36x^2\)

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

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

d) \(\left(x^2-3\right)^2+16=x^4-6x^2+25\)

\(=\left(x^4+10x^2+25\right)-16x^2\)

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

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

Y
17 tháng 6 2019

sửa câu b) xíu nha!

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

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

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

NV
17 tháng 6 2019

\(x^4-6x^2+25=x^4+10x^2+25-16x^2\)

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

Y
17 tháng 6 2019

\(=x^4-6x^2+25\)

\(=x^4+10x^2+25-16x^2\)

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

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

21 tháng 6 2019

\(\left(x^2-6x\right)^2-2\left(x-3\right)^2-81=\left[\left(x^2-6x\right)^2-81\right]-2\left(x-3\right)^2=\left[\left(x^2-6x\right)^2-9^2\right]-2\left(x-3\right)^2=\left(x^2-6x+9\right)\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x+11\right)\)

21 tháng 6 2019

=\(\left(x-3\right)^2\left(x^2-6x-11\right)\)

nha

1 tháng 5 2019

1) x2-2xy+y2-x+y
(=) (x-y)2-(x-y)
(=) [(x-y)-1].(x-y)
(=) (x-y-1).(x-y)
C= (x-y)(x2+xy+y2)-x(x2-y)+y(y2-x)
(=) x3-y3-x3+xy+y3-xy

(=)(x3-x3)+(-y3+y3)+(xy-xy)
(=) 0

1 tháng 5 2019

đúng nha

Y
17 tháng 6 2019

a) \(=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)

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

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

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

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

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

b) \(=x^4-x+2019\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)\

17 tháng 6 2019

c)\(x^4+2x^3+5x^2+4x-5\\=x^4+x^3+x^3-x^2+x^2+5x^2-x+5x-5\\ =x^2\left(x^2+x-1\right)+x\left(x^2+x-1\right)+5\left(x^2+x-1\right)=\left(x^2+x-1\right)\left(x^2+x+5\right)\)

14 tháng 4 2018

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

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

\(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x-1=-2;x=-1\\x-1=2;x=3\end{matrix}\right.\)

2 tháng 9 2019

Làm cho mk đi @Ender Dragon Boy Vcl

2 tháng 9 2019

(x−y+z)2+(z−y)2+2(x−y+z)(y−z)(x−y+z)2+(z−y)2+2(x−y+z)(y−z)

=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)=(x−y+z)2+(z−y)2+(x−y+z)(y−z)+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2+(x−y+z)(y−z)

=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)=(x−y+z)2+(x−y+z)(y−z)+(z−y)2−(x−y+z)(z−y)

=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]=(x−y+z)(x−y+y+z−z)+(z−y)[z−y−(x−y+z)]

=(x−y+z)x+(z−y)(z−y−x+y−z)=(x−y+z)x+(z−y)(z−y−x+y−z)

=x2−xy+xz+(z−y)(−x)=x2−xy+xz+(z−y)(−x)

=x2−xy+xz−xz+xy=x2−xy+xz−xz+xy

=x2

2 tháng 9 2019

\(x;y;z\rightarrow q;h;p\)

\(=\left(q^2+h^2+p^2\right)\left(q^2+h^2+p^2+2qh+2hp+2qp\right)+\left(qh+hp+pq\right)^2\)

\(Dat:\hept{\begin{cases}q^2+h^2+p^2=f\\qh+hp+qp=g\end{cases}}\Rightarrow\left(p^2+h^2+q^2\right)\left(p+q+h\right)^2+\left(qh+pq+ph\right)^2\)

\(=f\left(f+2g\right)+g^2=f^2+2fg+g^2=\left(f+g\right)^2=\left(q^2+h^2+p^2+qh+hp+pq\right)^2\)

2 tháng 9 2019

shitbo Cho đệ sửa lại bài SP chứ bài SP dài quá ạ:p

\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)

\(=\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2+2xy+yz+zx\right)+\left(xy+yz+zx\right)^2\)

Đặt \(x^2+y^2+z^2=a;xy+yz+zx=b\)

\(\Rightarrow a\left(a+2b\right)+b^2=a^2+2ab+b^2=\left(a+b\right)^2=\left(x^2+y^2+z^2+xy+yz+zx\right)^2\)

13 tháng 3 2021

Bạn tham khảo cách giải:

13 tháng 3 2021

Bạn tham khảo cách giải: