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a: Q=M+N

\(=5x^2y+5x+3-3xy^2z+xy^2z-4x^2y+5x-5\)

\(=x^2y+10x-2-2xy^2z\)

\(P=M-N\)

\(=5x^2y+5x+3-3xy^2z-xy^2z+4x^2y-5x+5\)

\(=9x^2y+8-4xy^2z\)

H=N-M

=-(M-N)

\(=-9x^2y-8+4xy^2z\)

b: \(Q=x^2y+10x-2-2xy^2z\)

=>Q có bậc là 4

\(P=9x^2y+8-4xy^2z\)

=>P có bậc là 4

\(H=-9x^2y-8+4xy^2z\)

=>H có bậc là 4

c: Khi x=-1;y=3;z=-2 thì

\(Q=\left(-1\right)^2\cdot3+10\cdot\left(-1\right)-2-2\cdot\left(-1\right)\cdot3^2\cdot\left(-2\right)\)

\(=3-10-2+2\cdot9\cdot\left(-2\right)\)

\(=-9-36=-45\)

Khi x=-1;y=3;z=-2 thì \(P=9\cdot\left(-1\right)^2\cdot3+8-4\cdot\left(-1\right)\cdot3^2\cdot\left(-2\right)\)

\(=27+8+4\cdot9\cdot\left(-2\right)\)

\(=35-72=-37\)

H=-P

=>H=37

a) P + Q = (x² + 2x³ - xy² + 5) + (x³ + xy² - 2x²y - 6)

= x² + 2x³ - xy² + 5 + x³ + xy² - 2x²y - 6

= (2x³ + x³) + x² + (-xy² + xy²) - 2x²y + (5 - 6)

= 3x³ + x² - 2x²y - 1

b) Q = P + N

N = Q - P

= (x³ + xy² - 2x²y - 6) - (x² + 2x³ - xy² + 5)

= x³ + xy² - 2x²y - 6 - x² - 2x³ + xy² - 5

= (x³ - 2x³) + (xy² + xy²) - 2x²y - x² + (-6 - 5)

= -x³ + 2xy² - 2x²y - x² - 11

Vậy N = -x³ + 2xy² - 2x²y - x² - 11

Lời giải:

a.

$A+B=(5x^2-7x+2)+(4x^2+3x-1)=9x^2-4x+1$
$A-B=(5x^2-7x+2)-(4x^2+3x-1)=x^2-10x+3$

b. 

$A(x)=2x^2-x+m=x(2x-5)+4x+m=x(2x-5)+2(2x-5)+m+10$

$=B(x)(x+2)+m+10$

Để $A(x)\vdots B(x)$ thì $m+10=0\Leftrightarrow m=-10$

bài 1

a) \(7x\left(5x-1\right)+5x-1=\left(5x-1\right)\left(7x+1\right)\)

b) \(4xy-4x^2-y^2+25=25-\left(4x^2-4xy+y^2\right)\)

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

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

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

d) \(3x^2-7x+2=3x^2-6x-x+2\)

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

\(=\left(x-2\right)\left(3x-1\right)\)

bài 2

a) * Rút gọn:

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

\(Q=\left[3\left(4x^2-4x+1\right)\right]+\left[2\left(2x^2-2x+3x-3\right)\right]-\left(x^2-9\right)\)

\(Q=\left(12x^2-12x+3\right)+\left(4x^2-4x+6x-6\right)-\left(x^2-9\right)\)

\(Q=12x^2-12x+3+4x^2-4x+6x-6-x^2+9\)

\(Q=15x^2-10x+6=5x\left(3x-2\right)+6\)

Thế x = 2 vào biểu thức Q ta được:

\(Q=5\cdot2\left(3\cdot2-2\right)+6=46\)

b) \(Q=5x\left(3x-2\right)+6=6\)

\(\Leftrightarrow5x\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)