phân tích thành nhân tử 1-m^2+2mn-n^2
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a: Ta có: \(m^2+2mn+n^2-p^2+2pq+q^2\)
\(=\left(m+n\right)^2-\left(p-q\right)^2\)
\(=\left(m+n-p+q\right)\left(m+n+p-q\right)\)
\(m^2-16+n^2-2mn\)
\(=n^2-2mn+m^2-16\)
\(=\left(n-m\right)^2-16\)
\(=\left(n-m-4\right)\left(n-m+4\right)\)
m2 - 16 + n2 - 2mn
= m2 - 2mn + n2 - 16
= (m - n)2 - 42
= (m - n - 4)(m - n + 4)
Easy \(x^2-n^2-2xy+y^2-m^2+2mn\)
\(=\left(x^2-2xy+y^2\right)-\left(n^2-2mn+m^2\right)\)
\(=\left(x-y\right)^2-\left(n-m\right)^2\)
\(=\left(x-y-n+m\right)\left(x-y+n-m\right)\)
\(x^2-n^2-2xy+y^2-m^2+2mn\)
\(=\left(x^2-2xy+y^2\right)-\left(n^2-2mn+m^2\right)\)
\(=\left(x-y\right)^2-\left(n-m\right)^2\)
\(=\left(x-y-n+m\right)\left(x-y+n-m\right)\)
làm a) thui nhé,b) theo đó mà làm
a) = (x-y)2 - (m-n)2 =(x-y +m-n)(x-y -m+n)
đơn giản như chơi game
b) \(a^2-10a+25-y^2-4yz-4z^2\)
\(=\left(a-5\right)^2-\left(y+2z\right)^2\)
\(=\left(a-5-y-2z\right)\left(a-5+y+2z\right)\)
\(=m^2-\left(n-2\right)^2=\left(m-n+2\right)\left(m+n-2\right)\)
\(=\left(m-n\right)\left(m+n\right)-2\left(m+n\right)=\left(m+n\right)\left(m-n-2\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(2\left(a-2b\right)+3a\left(2b-a\right)=2\left(a-2b\right)-3a\left(a-2b\right)=\left(2-3a\right)\left(a-2b\right)\)
\(a\left(m-n\right)-5\left(n-m\right)=a\left(m-n\right)+5\left(m-n\right)=\left(a+5\right)\left(m-n\right)\)
a) Ta có: \(2\left(a-2b\right)+3a\left(2b-a\right)\)
\(=2\left(a-2b\right)-3a\left(a-2b\right)\)
\(=\left(a-2b\right)\left(2-3a\right)\)
b) Ta có: \(a\left(m-n\right)-5\left(n-m\right)\)
\(=a\left(m-n\right)+5\left(m-n\right)\)
\(=\left(m-n\right)\left(a+5\right)\)