a: \(\dfrac{A}{B}=-\dfrac{1}{2}x^{m-2}y^{5-m}z\)

Để đây là phép chia hết thì m-2>=0 và 5-m>=0

=>2<=m<=5

b: \(\dfrac{A}{B}=-\dfrac{1}{2}xy^{m-1}z^{2-m}\)

Để đây là phép chia hết thì m-1>=0 và 2-m>=0

=>1<=m<=2

à....cái đó thì mình chưa tính ra được

mk chịu

khó quá

a ) \(x^3z+x^2yz-x^2z^2-xyz^2=\left(x^3z-x^2z^2\right)+\left(x^2yz-xyz^2\right)\)

\(=\left(x-z\right)\left(x^2z+xyz\right)\)

\(=xz\left(x-z\right)\left(x+y\right)\)

b ) \(p^{m+2}.q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)

\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)

\(=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)

\(=pq\left(p-q^2\right)\left(p^m-q^n\right)\)

\(a,x^2yz-x^3y^3z+xyz^2\)

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

\(b,4x^3+24x^2-12xy^2\)

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

\(c,15a^{m+2}b-45a^mb\)

\(=15a^m.a^2b-45a^mb\)

\(=15a^mb\left(a^2-3\right)\)

\(d,a^2-b^2+4bc-4c^2\)

\(=a^2-\left(b^2-4bc+4c^2\right)\)

\(=a^2-\left(b-2c\right)^2\)

\(=\left(a-b+2c\right)\left(a+b-2c\right)\)

a) \(x^2yz-x^3y^3z+xyz^2\)

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

b) \(4x^3+24x^2-12xy^2\)

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

c) \(15a^{m+2}.b-45a^m.b\)

\(=15.\left(a^m.a^2-3a^m.b\right)\)

\(=15.a^m.\left(a^2-3b\right)\)

d) \(a^2-b^2+4bc-4c^2\)

\(=a^2-\left(b^2-4bc+4c^2\right)\)

\(=a^2-\left[\left(b^2-2bc+c^2\right)-2bc+3c^2\right]\)

...... ;)))))))

a.\(\text{\(x^3z+x^2yz-x^2z^2-xyz^2\)}\)

\(=\left(x^3z+x^2yz\right)-\left(x^2z^2+xyz^2\right)=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)

\(=xz\left(x-y\right)\left(x+y\right)\)

\(\text{ }\)

b.gọi biểu thức là P ta có :

\(P=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)

\(P=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)=pq\left(p-q^2\right)\left(p^m-q^n\right)\)

aai làm được câu ni ko