Câu 1: A

Câu 21: A

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

1a

a: \(P=x^2-5x+6\)

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

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

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

b: \(P=3x^2+14x-5\)

\(=3x^2+15x-x-5\)

\(=3x\left(x+5\right)-\left(x+5\right)\)

\(=\left(x+5\right)\left(3x-1\right)\)

c: \(P=-2x^2-7x-5\)

\(=-\left(2x^2+7x+5\right)\)

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

\(=-\left[2x\left(x+1\right)+5\left(x+1\right)\right]\)

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

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

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

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

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

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)

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. (2x - 5)²

b. x(x² - 1) = x(x - 1)(x + 1)

c. (x - 3)(x² + 3x + 9)

d. 5x(x - y) + (x - y) = (x - y)(5x + 1)

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

=(x-2)(2+y)

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

c: =(x-7)(x+2)

a.

2x - 4 + xy - 2y

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

= (x-2)(y+2)

c.

x^2 - 5x - 14

= x^2 + 2x - 7x - 14

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

= (x-7)(x+2)

1. 158² - 116.158 + 58²

= 158² - 2.58.158 + 58²

= (158 - 58)²

= 100² = 10000

2.a. 8x² + 6x = 2x(4x + 3)

b, x³ - 5x² - 4x + 20

= x³ - 4x - 5x² + 20

= x(x² - 4) - 5(x² - 4)

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