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xz + yz - 5( x+y)

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

(x+y) (z-5)

Đề sai hả bạn?

\(2x^2-x^2-3x-1\)

\(=x^2-3x-1\)

\(=\frac{1}{4}\left(4x^2-12x-4\right)\)

\(=\frac{-1}{4}\left[13-\left(4x-12x+9\right)\right]\)

\(=-\frac{1}{4}\left[13-\left(2x-3\right)^2\right]\)

\(=-\frac{1}{4}\left(\sqrt{13}-2x+3\right)\left(\sqrt{13}+2x-3\right)\)

Nguyễn Công Tỉnh.Bad maths.^-^

Bài 1 :

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

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

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

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

Tick đúng nha 

X^2-6+8

x^7 + x^5 + 1
=x^7-x^6+x^5-x^3+x^2+x^6-x^5+x^4-x^2+x+x^5-x^4+x^3-x+1
=(x^2+x+1)(x^5-x^4+x^3-x+1)
+x^4 - 6x^3 + 12x^2 - 14x + 3
=x^4-2x^3+3x^2-4x^3-6x^2-12x+x^2-2x+3
=(x^2-4x+1)(x^2-2x+3)

 x^7 + x^5 + 1
=(x^2+x+1)(x^5-x^4+x^3-x+1)

\(x^7+x^5+1\)

\(=\left(x^7-x\right)\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

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

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

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

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