a) \(M=a^2\left(a+b\right)-b\left(a^2-b^2\right)+1=a^3+a^2b-a^2b+b^3+1=a^3+b^3+1\)

b) \(P=x\left(x-y+1\right)-y\left(y+1-x\right)-2=x^2-xy+x-y^2-y+xy-2=x^2+x-y-y^2-2\)

c) \(Q=\left(m+3\right)\left(m^2+3m-5\right)+\left(6-m\right)m^2+11=m^3+3m^2-5m+3m^2+9m-15+6m^2-m^3+11=12m^2+4m-4\)

a: Ta có: \(M=a^2\left(a+b\right)-b\left(a^2-b^2\right)+1\)

\(=a^3+a^2b-a^2b+b^3+1\)

\(=a^3+b^3+1\)

\(12+3y+4x+xy-12+2y+6x-x-5y\)

\(=9x+xy\)

\(=x\left(y+9\right)\)

12+3y+4x+xy-12+2y+6x-x-5y=

=9x+xy

=x(9+y)

12 + 3y + 4x + xy -12 + 2y + 6x- x - 5y

=9x +xy

=x(y+9)

Đề lỗi rồi kìa, bạn viết lại đi tridung

\(N=5x^2y+3x^2y+\dfrac{2}{3}x^2.12y\\ N=8x^2y+8x^2y\\ N=16x^2y\)

b: \(x-2\sqrt{xy}+y=\left(\sqrt{x}-\sqrt{y}\right)^2\)

\(a,2\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2+\left(x+1\right)^2\)

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

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

\(b,\left(x-y+1\right)^2+\left(1-y\right)^2+2\left(x-y+1\right)\left(y-1\right)\)

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

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

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

đề cuối phải sửa cái cuối thành \(\left(3x+5\right)^2\) 

\(c,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

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

Ta có :

[x-2]+[x-4]+[x-6]

=x-2+x-4+x-6

=(x+x+x)+(-2-4-6)

=3x-12

=3.(x-4)

câu hỏi trên có điều kiện đấy

x<5

Le Thi Thanh : bạn biểu mk thu gọn thôi

a)      Cách 1:

\(6(y - x) - 2(x - y)\)

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

\( = 8y - 8x\)

Cách 2:

\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)

b)      \(3{x^2} + x - 4x - 5{x^2}\)

\( = (3{x^2} - 5{x^2}) + (x - 4x)\)

\( =  - 2{x^2} - 3x\)