bang 68\2

 = 5/2x^2y-5/3x-2y

k mk nha

5/2x^2y-5/3x-2y là sao bạn

1) \(4x^5y^2-8x^4y^2+4x^3y^2\)

\(=4x^3y^2\left(x^2-2x+1\right)\)

\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=4x^3y^2\left(x-1\right)^2\)

2) \(5x^4y^2-10x^3y^2+5x^2y^2\)

\(=5x^2y^2\left(x^2-2x+1\right)\)

\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=5x^2y^2\left(x-1\right)^2\)

3) \(12x^2-12xy+3y^2\)

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

\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

4) \(8x^3-8x^2y+2xy^2\)

\(=2x\left(4x^2-4xy+y^2\right)\)

\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

5) \(20x^4y^2-20x^3y^3+5x^2y^4\)

\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)

\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

\(=5x^2y^2\left(2x-y\right)^2\)

1: 4x^5y^2-8x^4y^2+4x^3y^2

=4x^3y^2(x^2-2x+1)

=4x^3y^2(x-1)^2

2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)

3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)

4: \(=2x\left(4x^2-4xy+y^2\right)=2x\left(2x-y\right)^2\)

5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)

aaaaaaaaaaa

\(\frac{2x^4y^3+6x^3y^2-10x^2y}{-2x^2y}\)\(=\frac{2x^2y\left(x^2y^2+3xy-5\right)}{-2x^2y}\)\(=5-3xy-x^2y^2\)

Lời giải:

a) $P(x)= 5x+x^3y-2xy+4x^3y+3x^2y-10x$

$=(x^3y+4x^3y)+3x^2y-2xy+(5x-10x)$

$=5x^3y+3x^2y-2xy-5x$

$Q(x)=4x-5x^3y+2x^2y-x^3y+6xy+11x^3-8x$

$=-6x^3y+2x^2y+11x^3+6xy-4x$

$P(x)-Q(x)=11x^3y+x^2y-8xy-x-11x^3$

Bậc của $P(x)-Q(x)$ là $3+1=4$

b)

$P(x)+Q(x)=-x^3y+5x^2y+4xy-9x+11x^3$

$P(x)-Q(x)$ đã thu gọn ở phần a.

\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)

\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)

\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)

\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)

a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)

c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)