Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

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

=(x^2+x)^2+3(x^2+x)-10

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

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

2: \(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

\(=\left(x^2+5ax\right)^2+10a^2\left(x^2+5ax\right)+25a^2\)

\(=\left(x^2+5ax+5a^2\right)^2\)

3: \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

5: \(M=\left(n+1\right)\left(n^2+2n\right)+360\)

=n(n+1)(n+2)+360 chia hết cho 6

6A

7D

\(x^3-x-12\)

\(=x^3-4x+3x-12\)

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

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

\(=\left(x-4\right)\left[x\left(x+4\right)+3\right]\)

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

x² - 2xy - xz + yz + y² 

= (x² - 2xy + y²) - (xz - yz)

= (x - y)² - z . (x - y)

= (x - y)(x - y - z) 

x³ +2x²y-x-2y

= (x³ - x) + (2x²y - 2y)

= x(x² - 1) + 2y(x² - 1)

= (x² - 1) (x + 2y)

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

a)

\(6xy+12x-4y-8\\ =2\left(3xy+6x-2y-4\right)\\ =2\left[3x\left(y+2\right)-2\left(y+2\right)\right]\\ =2\left(y+2\right)\left(3x-2\right)\)

b)

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

\(a,6xy+12x-4y-8=\left(6xy+12x\right)-\left(4y+8\right)=6x\left(y+2\right)-4\left(y+2\right)=\left(y+2\right).\left(6x-4\right)=\left(y+2\right).2.\left(3x-2\right)\)

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

a)

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

b)

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

=9x^2+6x+1-x^2-x+5

=8x^2+5x+6

\(\left(3x+1\right)^2-\left(x^2+x-5\right)\\ =9x^2+6x+1-x^2-x+5\)

\(=9x^2-x^2+6x-x+1+5\)

\(=8x^2+5x+6\)

=9x^2 +6x +1 -x^2 -x +5

=8x^2 +5x +6

\(b)3x(x+5)-2x-10=0\)

\(<=>3x(x+5)-2(x+5)=0\)

\(<=>(x+5)(3x-2)=0\) 

\(<=>\) \(\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)

a: =>x^2+8x+16-x^2+2x-3x+6=-13

=>7x+22=-13

=>7x=-35

=>x=-5

b: =>(x+5)(3x-2)=0

=>x=2/3 hoặc x=-5

a.

\(\Leftrightarrow x^2+8x+16-\left(x^2+x-6\right)=-13\)

\(\Leftrightarrow x^2+8x+16-x^2-x+6=-13\)

\(\Leftrightarrow7x=-35\)

\(\Leftrightarrow x=-5\)

b.

\(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)