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a, \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b, \(4m^2-16n^2=\left(2m-4n\right)\left(2m+4n\right)=4\left(m-2n\right)\left(m+2n\right)\)
c, \(49-16x^2=\left(7-4x\right)\left(7+4x\right)\)
d, \(25-9y^2=\left(5-3y\right)\left(5+3y\right)\)
e, \(81x^2-16y^2=\left(9x-4y\right)\left(9x+4y\right)\)
\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)
\(=3\left(a-b+c\right)\left(x+6y\right)^2\)
___________________
\(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(n^2\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)n\)
Ta có:
\(\left\{{}\begin{matrix}\left(n-1\right).n.\left(n-1\right)\text{⋮}3\\\left(n-1\right)n\text{⋮}2\\\left(n+1\right)n\text{⋮}2\end{matrix}\right.\)
⇒ \(n\left(n-1\right)n\left(n+1\right)\text{⋮}2.2.3=12\)
a: \(A=2\left(m^3+n^3\right)-3\left(m^2+n^2\right)\)
\(=2\left[\left(m+n\right)^3-3mn\left(m+n\right)\right]-3\left[\left(m+n\right)^2-2mn\right]\)
\(=2-6mn-3+6mn\)
=-1
c: \(C=\left(a-1\right)^3-4a\left(a+1\right)\left(a-1\right)+3\left(a-1\right)\left(a^2+a+1\right)\)
\(=a^3-3a^2+3a-1-4a\left(a^2-1\right)+3a^3-3\)
\(=4a^3-3a^2+3a-4-4a^3+4a\)
\(=-3a^2+7a-4\)
\(=-3\cdot9-21-4\)
=-27-21-4
=-52
Phân tích thành nhân tử đúng không ạ
a, (m+n)(m-n)
b, 4(m-2n)(m+2n)
a_ $m^2-n^2 = (m-n)(m+n)$
b_ $4m^2-16n^2$
$ = (2m)^2 - (4n)^2$
$ = (2m-4n)(2m+4n)$
$ = 4.(m-2n)(m+2n)$