a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

\(=\left(x^2-16\right)\left(x^2+2\right)\)

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

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

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

Bài 3: 

b: \(x^2+2x+1=\left(x+1\right)^2\)

c: \(x^2-16=\left(x-4\right)\left(x+4\right)\)

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

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

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

B1

A=11x^2-x-2

B=2(-4+x)

B2

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

B2

b)=(5x-7)(x+2)

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

b) \(5x\left(x-3\right)+\left(3-x\right)^2-\left(x-3\right)=5x\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)=\left(x-3\right)\left(5x+x-3-1\right)=\left(x-3\right)\left(6x-4\right)=2\left(x-3\right)\left(3x-2\right)\)

c) \(9x^2y^3-3x^4y^2-6x^3y^2+16xy^4=xy^2\left(9xy-3x^3-6x^2+16y^2\right)\)

Câu 1: A

Câu 21: A

\(16,A\\ 17,C\\ 18,A\\ 19,C\\ 20,A\\ 21,A\)

1a

Bài 2

a) 5x² + 30y

= 5(x² + 6y)

b) x³ - 2x² - 4xy² + x

= x(x² - 2x - 4y² + 1)

= x[(x² - 2x + 1) - 4y²]

= x[(x - 1)² - (2y)²]

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

Bài 3:

a: \(2x\left(x-3\right)-x+3=0\)

=>\(2x\left(x-3\right)-\left(x-3\right)=0\)

=>(x-3)(2x-1)=0

=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

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

=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)

=>\(5x^2-x-2=5x^2\)

=>-x-2=0

=>-x=2

=>x=-2

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

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

a: \(4x^2-12x+9-3\left(2x-3\right)\left(x+1\right)\)

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

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

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

b: \(25-4x^2+8xy-4y^2\)

\(=25-\left(2x-2y\right)^2\)

\(=\left(5-2x+2y\right)\left(5+2x-2y\right)\)

1.

a) \(2x^4-4x^3+2x^2\)

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

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

b) \(2x^2-2xy+5x-5y\)

\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)

\(=2x\left(x-y\right)+5\left(x-y\right)\)

\(=\left(x-y\right)\cdot\left(2x+5\right)\)

2 . 

a,

\(4x\left(x-3\right)-x+3=0\)

⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)

⇒\(\left(x-3\right)\left(4x-1\right)=0\)

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

⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)

vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)

b, 

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

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

⇒\(\left(x-4\right)\left(3x-2\right)=0\)

⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)

ban tích cho mk vs nha

a: (3x-5)^2-(x+3)^2

=(3x-5-x-3)(3x-5+x+3)

=(2x-8)(4x-2)

=2(x-4)*2*(2x-1)

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

b: (2x+1)^2-4(x-3)^2

=(2x+1)^2-[2*(x-3)]^2

=(2x+1)^2-(2x-6)^2

=(2x+1-2x+6)(2x+1+2x-6)

=(4x-5)*7