Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Easy \(x^2-n^2-2xy+y^2-m^2+2mn\)
\(=\left(x^2-2xy+y^2\right)-\left(n^2-2mn+m^2\right)\)
\(=\left(x-y\right)^2-\left(n-m\right)^2\)
\(=\left(x-y-n+m\right)\left(x-y+n-m\right)\)
\(x^2-n^2-2xy+y^2-m^2+2mn\)
\(=\left(x^2-2xy+y^2\right)-\left(n^2-2mn+m^2\right)\)
\(=\left(x-y\right)^2-\left(n-m\right)^2\)
\(=\left(x-y-n+m\right)\left(x-y+n-m\right)\)
\(m^2-16+n^2-2mn\)
\(=n^2-2mn+m^2-16\)
\(=\left(n-m\right)^2-16\)
\(=\left(n-m-4\right)\left(n-m+4\right)\)
m2 - 16 + n2 - 2mn
= m2 - 2mn + n2 - 16
= (m - n)2 - 42
= (m - n - 4)(m - n + 4)
Phân tích thành nhân tử
\(=\left(my+nx\right)\left(ny+mx\right)\)
mn(x2 +y2) +xy(m2 +n2)= mnx2 +mny2 +xym2 +xyn2
=mx(nx + my) +ny( my +nx)
=(mx+ny)(nx+my)
a: \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)
\(=xz\left(x+y\right)\left(x-z\right)\)
A = (n2 - 15)2 + 64
A = n4 - 30n2 + 225 + 64
A = n4 - 30n2 + 289
A = (n2 -15)2 + 64
= (n2- 15)2 + 82
= (n2 - 15 + 8).(n2 - 15 - 8)
=(n2 -7).(n2 - 23)
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: Ta có: \(m^2+2mn+n^2-p^2+2pq+q^2\)
\(=\left(m+n\right)^2-\left(p-q\right)^2\)
\(=\left(m+n-p+q\right)\left(m+n+p-q\right)\)