Biểu thức này không phân tích được thành nhân tử nhé. Bạn xem xem có viết sai đề không.

= (3x + 1 - x - 1)(3x + 1 + x + 1)

= 2x(4x + 2)

Em áp dụng hđt số 3 trong sgk nhé.

(3x+1+x+1) (3x+1-x-1) = (4x+2) 2x

                                      = 4x(2x+1)

o: x^4+x^3+x^2-1

=x^3(x+1)+(x-1)(x+1)

=(x+1)(x^3+x-1)

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

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

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

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

s: =(2xy)^2-(x^2+y^2-1)^2

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

=[1-(x^2-2xy+y^2]+[(x+y)^2-1]

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

u: =(x^2-y^2)-4(x+y)

=(x+y)(x-y)-4(x+y)

=(x+y)(x-y-4)

x: =(x^3-y^3)-(3x-3y)

=(x-y)(x^2+xy+y^2)-3(x-y)

=(x-y)(x^2+xy+y^2-3)

z: =3(x-y)+(x^2-2xy+y^2)

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

=(x-y)(x-y+3)

o) \(x^4+x^3+x^2-1\)

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

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

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

q) \(x^3+x^2y-xy^2-y^3\)

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

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

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

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

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

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

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

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

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

u) \(x^2-y^2-4x-4y\)

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

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

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

x) \(x^3-y^3-3x+3y\)

\(=\left(x^3-y^3\right)-\left(3x-3y\right)\)

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

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

z) \(3x-3y+x^2-2xy+y^2\)

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

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

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

c:   x^6 - y^6 

= (x^3)^2 - (y^3)^2 

= (x^3 - y^3) (x^3 + y^3)

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

 a:   x^2 + 4y^2 + 4xy

= x^2 + 4xy + 4y^2

= (x+2y)^2

Mình cảm ơn nha. Bạn giải đc ý b k ạ

(y²+y)+(-xy-x) = y(y+1)-x(y+1) = (y-x)(y+1)

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

\(x^3-x^2y+3x-3y\)

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

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

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

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

3x^2-3x-6

=3(x^2-x-2)

=3(x-2)(x+1)