=(2x+3)^{2 -} (2x+1)²

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

(2x+3) (2x-3)- (2x+1)²

= (2x)^2 - 3^2 - (2x+1)²

= 4x^2 - 9 - (2x+1)²

= 4x^2 -9 - 4x² - 4x -1

=(4x² - 4x²) - (9+1) - 4x

= -10 -4x

địt ko em

\(A=\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27\)

\(A=\left(x^2-3x-1\right)^2-2\left(x^2-3x-1\right)6+36-9\)

\(A=\left(x^2-3x-1-6\right)^2-3^2\)

\(A=\left(x^2-3x-7\right)^2-3^2\)

\(A=\left(x^2-3x-7-3\right)\left(x^2-3x-7+3\right)\)

\(A=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

( 2x + 3 )( 2x - 3 ) - ( 2x + 1 )²

= ( 2x )² - 3² - ( 4x² + 4x + 1 )

= 4x² - 9 - 4x² - 4x - 1 = -4x - 10

=3x²-6x-x+2 

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

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

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

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

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

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

1,=(x+4)(x-4)

2,=(2a-1)(2a+1)

4,=(5-3y)(5+3y)

5,=(a+1-4)(a+1+4)

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

7,=(a+b-a+b)(a+b+a-b)

8,=(a+x)²

9,=(x-2)²

10,=(x-3y)²

11,=(x+2)(x²-2x+4)

12,=(a+3)(a²-3a+9)

13,=(3x-1)(9x²+9x+1)

14,=(1/2-b)(1/4+1/2b+b²)

15,=(2x-x+y)(2x+x+y)

hello Đam Mê Ngành Luật:) sửa +3 thành -3

\(4x^2+9y^2-12xy+4x-6y-3\)

\(=\left(4x^2-12xy+9y^2+4x-6y+1\right)-4\)

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

Đề vẫn đúng là +3 bae à :<< cậu xem lại xem

\(1,x^3-2x+x\)

\(x^3-x\)

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

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

\(2,3x^2+6x+3-3y^2\)

\(\left(\sqrt{3}x+\sqrt{3}\right)^2-\left(\sqrt{3}y\right)^2\)

\(\left(\sqrt{3}x+\sqrt{3}-\sqrt{3}y\right)\left(\sqrt{3}x+\sqrt{3}+\sqrt{3}y\right)\)

\(3.\left(x+1-y\right)\left(x+1+y\right)\)

\(3,2xy-x^2-y^2+16\)

\(-\left(x^2-2xy+y^2-16\right)\)

\(-\left[\left(x-y\right)^2-4^2\right]\)

\(-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)

gọi chiều dài quãng đường là x km

khi đó thời gian đi từ A đến B là :\(\frac{x}{60}\text{ giờ}\)

thời gian trở về A là :\(\frac{x-10}{50}\text{ giờ}\)

ta có phương trình : \(\frac{x-10}{50}-\frac{x}{60}=1\Leftrightarrow60x-600-50x=3000\Leftrightarrow x=360\left(km\right)\)

ta có 

\(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)

\(=\left[\left(x^2+10x+12\right)-12\right]\left[\left(x^2+10x+12\right)+12\right]+128\)

\(=\left(x^2+10x+12\right)^2-12^2+128=\left(x^2+10x+12\right)^2-16\)

\(=\left(x^2+10x+12-4\right)\left(x^2+10x+12+4\right)=\left(x^2+10x+8\right)\left(x^2+10x+16\right)\)

\(=\left(x+2\right)\left(x+8\right)\left(x^2+10x+8\right)\)