Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp

\(xy\left(x+y\right)-yz\left(y+z\right)-zx\left(x+z\right)\)

\(=x^2y+xy^2-y^2z-yz^2-zx^2-z^2x\) 

\(x^4-y^4+2x^3y-2xy^3\)

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

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

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

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

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

`a, 2x^2+3x-5=0`

`<=> 2x^2+5x-2x-5=0`

`<=> x(2x+5) -(2x+5)=0`

`<=> (2x+5)(x-1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-5\\x=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

`b,-4x^3+x^2-8x+2=0`

`<=>- x^2(4x-1)-(8x-2)=0`

`<=> -x^2(4x-1) -2(4x-1)=0`

`<=>(4x-1)(-x^2-2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\-x^2-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=1\\-x^2=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x\in\varnothing\end{matrix}\right.\)

`c, 36x^3-4x=0`

`<=>4x(9x^2-1)=0`

`<=>4x(3x-1)(3x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=1\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

ax - bx - a² + 2ab - b²

= (ax - bx) - (a² - 2ab + b²)

= x(a - b) - (a - b)²

= (a - b)(x - a + b)

\(3x^2-7xy+2y^2\)

\(=3x^2-6xy-xy+2y^2\)

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

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

\(x^2-y^2+10x-10y\)

\(=\left(x^2-y^2\right)+\left(10x-10y\right)\)

\(=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+10\right)\)

x(x+1)(x+2)(x+3)+1

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

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

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

Để giải phương trình (4-2)^2 - 16 = 0, ta thực hiện các bước sau:

(4-2)^2 - 16 = 0

(2)^2 - 16 = 0

4 - 16 = 0

-12 = 0

Phương trình trên không có nghiệm vì -12 không bằng 0.

Em xem lại đề nhé. Đề sai rồi, bổ sung đề cho đầy đủ

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

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

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

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

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