\(16x-5x^2-3=-\left(5x^2-16x+3\right)=-\left[\left(5x^2-15x\right)-\left(x-3\right)\right]=-\left[5x\left(x-3\right)-\left(x-3\right)\right]=-\left(5x-1\right)\left(x-3\right)=\left(5x-1\right)\left(3-x\right)\)

16x - 5x² - 3 = -5x² + 16x - 3 = -5x² + 15x + x - 3

= -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x) 

 12x^3+16x^2-5x-3 = 12x³+18x² -2x²-3x -2x -3

= 12x²(x +3/2) -2x(x+3/2) -2(x+3/2)

=(12x²-2x-2)(x+3/2)

=2(6x²-x-1)(x+3/2)

=2[6x(x+1/3)-3(x+1/3)](x+3/4)

=2(6x-3)(x+1/3)(x+3/4)

=6(2x-1)(x+1/3)(x+3/4)

k cho mk  nha

\(-5x^2+16x-3\)

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

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

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

16x - 5x^2 - 3

=x - 5x^2 + 15x - 3

=x(1-5x) + 3(5x-1)

=3(5x-1) -x(5x-1)

=(3-x)(5x-1)

\(-5x^2+16x-3=\left(-5x^2+15x\right)+\left(x-3\right)\)

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

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

áp dụng hằng đẳng thức 7 và 3 nha dễ dàng mà . bn tách hết cho về mũ 3 hoặc mũ 2 nha 

a, 5x^2 +5xy - x - y 

= 5x ( x+ y ) - (x + y)

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

b, 2x^2  + 3x - 5 

= 2x^2 - 2x + 5x - 5 

= 2x( x - 1) + 5( x - 1)

= ( 2x + 5 )(x- 1 )

c; 16x - 5x^2 - 3 

c, = - ( 5x^2 - 16x + 3 )

   = - ( 5x^2 - x - 15x + 3 )

= - [ x(5x - 1 ) - 3 (5x - 1) ]

= - ( x- 3)(5x -  1 )

\(-5x^2+16x-3=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(-5x+1\right)\)

Phần 1 ) Đánh lộn dấu " ) " thành số 0 rồi bạn .... 

\(\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x+4\right)-15\)

\(=\left[\left(x+1\right).\left(x+4\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]-15\)

\(=\left(x^2+5x+4\right).\left(x^2+5x+6\right)-15\)

\(\text{Đặt }k=x^2+5x+4,\text{ta có: }\)

\(\left(x^2+5x+4\right).\left(x^2+5x+6\right)-15=k.\left(k+2\right)-15=k^2+2k-15=k^2-5k+3k-15=k.\left(k-5\right)+3.\left(k-5\right)=\left(k-5\right).\left(k+3\right)\)

tự thay vô nha : )