Phân tích đa thức thành nhân tử :
a ) \(x^3z+x^2yz-x^2z^2-xyz^2\) b ) \(p^{m+2}q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
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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 :
\(P=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)
\(P=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)=pq\left(p-q^2\right)\left(p^m-q^n\right)\)
a: \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)
\(=xz\left(x+y\right)\left(x-z\right)\)
Bài 3:
a: =>(2x-7)(x-2)=0
=>x=7/2 hoặc x=2
b: =>(x-1)(x+2)=0
=>x=1 hoặc x=-2
d: =>2x+3=0
hay x=-3/2
\(a,x^2yz-x^3y^3z+xyz^2\)
\(=xyz\left(x-x^2y^2+z\right)\)
\(b,4x^3+24x^2-12xy^2\)
\(=4\left(x^3+6x^2-3xy^2\right)\)
\(c,15a^{m+2}b-45a^mb\)
\(=15a^m.a^2b-45a^mb\)
\(=15a^mb\left(a^2-3\right)\)
\(d,a^2-b^2+4bc-4c^2\)
\(=a^2-\left(b^2-4bc+4c^2\right)\)
\(=a^2-\left(b-2c\right)^2\)
\(=\left(a-b+2c\right)\left(a+b-2c\right)\)
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) \(a^2-b^2+4bc-4c^2\)
\(=a^2-\left(b^2-4bc+4c^2\right)\)
\(=a^2-\left[\left(b^2-2bc+c^2\right)-2bc+3c^2\right]\)
...... ;)))))))
\(a,x^2-xy+9x-9y\)
\(=x\left(x-y\right)+9\left(x-y\right)\)
\(=\left(x+9\right)\left(x-y\right)\)
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 ) \(p^{m+2}.q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)
\(=pq\left(p-q^2\right)\left(p^m-q^n\right)\)