Ta có

4 x 2 y − 2 x y 2 + 1 3 x 2 y − x + 2 x 2 y + x y 2 − 1 3 x − 6 x 2 y = 4 x 2 y + 1 3 x 2 y + 2 x 2 y − 6 x 2 y + − 2 x y 2 + x y 2 + − 1 3 x − x = 1 3 x 2 y − x y 2 − 4 3 x

Chọn đáp án B

Chọn A

Ta có: 4x2y + (-3x2y) + 3x2y + 2x2y = 6x2y.

\(12x^4y^2\)                            \(12x^2y^3\)

\(-12x^3y^2\)                           \(12x^3y^2\)

x²y + 2x²y + 3x²y + 4x²y + .... + nx²y = 55x²y

\(\Rightarrow x^2y\left(1+2+...+n\right)=55x^2y\)

\(\Rightarrow1+2+...+n=55\)

\(\Rightarrow\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=55\)

\(\Rightarrow n\left(n+1\right)=55.2=110\)

\(\Rightarrow n^2+n-110=0\)

\(\Rightarrow n^2-10n+11n-110=0\)

\(\Rightarrow n\left(n-10\right)+11\left(n-10\right)=0\)

\(\Rightarrow\left(n-10\right)\left(n+11\right)=0\)

\(\Rightarrow n-10=0\) hay \(n+11=0\)

\(\Rightarrow n=10\left(nhận\right)\) hay \(n=-11\left(loại\right)\)

\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(a,=x\left(x-2\right)\\ b,=2b\left(x-3y\right)+a\left(x-3y\right)=\left(a+2b\right)\left(x-3y\right)\\ c,=x\left(x^2+2xy+y^2-4\right)=x\left[\left(x+y\right)^2-4\right]=x\left(x+y+2\right)\left(x+y-2\right)\\ d,=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\\ đ,=5\left(x-y\right)\left(x+y\right)+3\left(x+y\right)^2=\left(x+y\right)\left(5x-5y+3x+3y\right)\\ =\left(x+y\right)\left(8x-2y\right)=2\left(4x-y\right)\left(x+y\right)\\ e,=3x\left(2xy-3\right)\\ b,=x\left(4x^2-4xy+y^2-4\right)=x\left[\left(2x-y\right)^2-4\right]=x\left(2x-y-2\right)\left(2x-y+2\right)\\ f,=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

a,=x(x−2)b,=2b(x−3y)+a(x−3y)=(a+2b)(x−3y)c,=x(x2+2xy+y2−4)=x[(x+y)2−4]=x(x+y+2)(x+y−2)d,=4−(x+y)2=(2−x−y)(2+x+y)đ,=5(x−y)(x+y)+3(x+y)2=(x+y)(5x−5y+3x+3y)=(x+y)(8x−2y)=2(4x−y)(x+y)e,=3x(2xy−3)b,=x(4x2−4xy+y2−4)=x[(2x−y)2−4]=x(2x−y−2)(2x−y+2)f,=(x+y)2−z2=(x+y−z)(x+y+z)

\(4x^2y+7x^2y-6x^2y-3x^2y\)

\(=\left(4+7-6-3\right)x^2y\)

\(=2x^2y\)

$4{}^{}y+7{}^{}y-6{}^{}y-3{}^{}y$

$=\left(4+7-6-3\right){}^{}y$

$=2{}^{}y$

Thay x = 2, y = -3 vào đơn thức ta có A = -(2.2)2.(-3) = 24. Chọn C