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\(\left(x^2-1\right)^2-x\left(x^2-1\right)-2x^2\)
= \(\left(x^2-1\right)^2-2.\left(x^2-1\right).\frac{x}{2}+\frac{x^2}{4}-\frac{x^2}{4}-2x^2\)
= \(\left(x^2-1-\frac{x}{2}\right)^2-\frac{9}{4}x^2\)
\(=\left(x^2-1-\frac{x}{2}-\frac{3}{2}x\right)\left(x^2-1-\frac{x}{2}+\frac{3}{2}x\right)\)
= \(\left(x^2-2x-1\right)\left(x^2-x-1\right)\)
Phân tích tiếp được đấy:
\(x^2-2x-1=\left(x-1\right)^2-2=\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)\)
\(x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=\left(x-\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}+\frac{\sqrt{5}}{2}\right)\)
Thay vào nhé!
x^7+x^2+2
=(x^7+x^6+x^5)-(x^6+x^5+x^4)+(x^4+x^3+x^2) +(1 -x^3)
=x^5(x^2+1)-x^4(x^2+1)+x^2(x^2+1)+(1-x)(1+x+x^2)
=(x^2+1)(x^5-x^4+x^2-x+1)
\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2\)
\(=x^4+1+2x^2+3x^2+3x+2x^2\)
\(=x^4+3x^3+4x^2+3x+1\)
\(=x^4+x^3+2x^3+2x^2+2x^2+2x+x+1\)
\(=\left(x+1\right)\left(x^3+2x^2+2x+1\right)\)
Đặt \(x^2+1=a\) thay vào ta được :
\(a^2+3ax+2x^2\)
\(=a^2+ax+2ax+2x^2\)
\(=a\left(a+x\right)+2x\left(a+x\right)\)
\(=\left(a+2x\right)\left(a+x\right)\)
\(=\left(x^2+1+2x\right)\left(x^2+1+x\right)\)
\(=\left(x+1\right)^2\left(x^2+x+1\right)\)
\(a,x^2\left(1-x^2\right)-4-4x^2\)
\(=x^2-x^4-4-4x^2\)
\(=x^2-\left(x^4+4x^2+4\right)\)
\(=x^2-\left(x^2+2\right)^2\)
\(=\left(2x^2+2\right).\left(-2\right)\)
\(=-4\left(x^2+1\right)\)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
1, \(a^2b-2ab^2+ab\) = \(ab\left(a-2b+1\right)\)
2, \(a\left(x-1\right)+b\left(1-x\right)\)=\(a\left(x-1\right)-b\left(x-1\right)\)
=\(\left(x-1\right)\left(a-b\right)\)
a, 3x2 - 7x +2 = 3x2 - 6x - x +2 = 3x(x-2) - (x-2) = (3x-1)(x-2);
b, a(x2+1) -x(a2+1) = ax2+a-xa2-x=(ax2-xa2) -(x-a)=ax(x-a)-(x-a)=(ax-1)(x-a)
a) \(3x^2-7x+2=3x^2-3x-4x+2=3x\left(x^2-1\right)-2\left(x-1\right)\)
\(=3x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)=\left(x-1\right)\left[\left(3x.\left(x+1\right)-2\right)\right]\)
\(=\left(x-1\right)\left(3x^2+3x-2\right)\)
b) \(a\left(x^2+1\right)-x\left(a^2+1\right)=ax^2+a-xa^2-x\)
\(=ax\left(x-a\right)+\left(a-x\right)=ax\left(x-a\right)-\left(x-a\right)=\left(x-a\right)\left(ax-1\right)\)
(Mình chắc chắn là mình làm đúng, mong bạn ủng hộ và click cho mình nha nha!)
\(a\left(x^2+1\right)-x\left(a^2+1\right)=ax^2+a-xa^2-x=ax\left(x-a\right)-\left(x-a\right)=\left(ax-1\right)\left(x-a\right)\)