a) 4 x 2  + 12x + 9.                           b) 36 – 36u – 9 u 2 .

c) y 2  – 16.                                         d) a 2 4 - 4 a + 16.

a, - \(\dfrac{1}{3}\).\(xy\).(3\(x^3\).y² - 6\(x^2\) + y²)

= - \(x^4\).y³ + 2\(x^3\).y - \(\dfrac{1}{3}\).\(xy^3\)

b, (2\(x\) -3).(4\(x\)² + 6\(x\) + 9)

 = (2\(x\))³ - 3³

= 8\(x^3\) - 27 

\(\dfrac{2x}{x-y}\) - \(\dfrac{2y}{x-y}\) = \(\dfrac{2x-2y}{x-y}\) = \(\dfrac{2(x-y)}{x-y}\) = 2

1-3+5-7+...+21-23
=(1-3)+(5-7)+...+(21-23)
Tổng số cặp là:
[(23-1):2+1]:2=6
Nháp: [(23-1):2+1]:2=(22:2+1):2=12:2=6
=1-3=-2=-2*6=-12
Đây nha bạn.
Chúc bạn học tốt!

\(1-3+5-7+...+21-23\)

\(=\left(1+5+9+13+17+21\right)-\left(3+7+11+15+19+23\right)\)

\(=\left[\left(21-1\right):2+1\right].\left(21+1\right):2-\left[\left(23-3\right):2+1\right].\left(23+3\right):2\)

\(=11.11-11.13\)

\(=121-143\)

\(=-22\)

b: (3x-2)^5+(5-x)^5+(-2x-3)^5=0

Đặt a=3x-2; b=-2x-3

Pt sẽ trở thành:

a^5+b^5-(a+b)^5=0

=>a^5+b^5-(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5)=0

=>-5a^4b-10a^3b^2-10a^2b^3-5ab^4=0

=>-5a^4b-5ab^4-10a^3b^2-10a^2b^3=0

=>-5ab(a^3+b^3)-10a^2b^2(a+b)=0

=>-5ab(a+b)(a^2-ab+b^2)-10a^2b^2(a+b)=0

=>-5ab(a+b)(a^2-ab+b^2+2ab)=0

=>-5ab(a+b)(a^2+b^2+ab)=0

=>ab(a+b)=0

=>(3x-2)(-2x-3)(5-x)=0

=>\(x\in\left\{\dfrac{2}{3};-\dfrac{3}{2};5\right\}\)