4x² - 4y²

= (2x)² - (2y)²

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

(2x - 2y)(2x + 2y)

4x²-4y²=4.(x²-y²)=4.(x-y).(x+y)

$x(x+y)+4x+4y$

$=x(x+y)+4(x+y)$

$=(x+y)(x+4)$

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

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

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

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

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

 bÀI LÀM

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

Câu 1:

\(4x^2+16x-9\)

\(=4x^2+18x-2x-9\)

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

\(=\left(2x-1\right)\left(2x+9\right)\)

Câu 2:

\(6x^2-11x+3=0\)

\(\Leftrightarrow6x^2-2x-9x+3=0\)

\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)

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

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

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

\(=4x\left(x+2y\right)+3\left(x+2y\right)=\left(x+2y\right)\left(4x+3\right)\)

\(4x^2+8xy+3x+6y\)

\(=4\left(x+2y\right)+3\left(x+2y\right)\)

\(=7\left(x+2y\right)\)

=4x(x+2y)+3(x+2y)=(x+2y)(4x+3)

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

Phân tích thành nhân tử

\(4x^2-y^2+4x+1\)

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

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

= (4x² + 4x + 1) - y²

= (2x +1)² - y²

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