a^3m+2a^2m+a^m = a^m(a^2m+2a^m+1) = a^m[(a^m)²+2a^m+1] = a^m(a^m+1)²

Ta có :

a^3m+2a^2m+a^m 

= a^m(a^2m+2a^m+1)

= a^m[(a^m)²+2a^m+1]

= a^m(a^m+1)²

\(x^2-y^2+10x-10y=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)=\left(x-y\right)\left(x+y+10\right)\)

b) x^8+x^4+1

=x^8-x^2+x^4-x+x^2+x+1

=x^2(x^6-1)+x(x^3-1)+(x^2+x+1)

=x^2[(x^3)^2-1]+x(x^3-1)+(x^2+x+1)

=x^2(x^3-1)(x^3+1)+x(x^3-1)+(x^2+x+1)

=x^2(x-1)(x^2+x+1)(x^3+1)+x(x^3-1)+(x^2+x+1)

=x^2(x-1)(x^2+x+1)(x^3+1)+x(x-1)(x^2+x+1)+(x^2+x+1)

=(x^2+x+1)[x^2(x-1)(x^3+1)+x(x-1)+1]

=(x^2+x+1)(x^6+x^3-x^5-x+1)

dung thi tick cho minh nha minh thu may tinh roi

\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a\left(a ^3+3.a^22b+3.a2b^2+2b^3\right)-b\left(2a^3+3.2a^2.b+3.2a.b^2+b^3\right)\)

\(=a\left(a^3+6a^2b+6ab^2+8b^3\right)-b\left(8a^3+6a^2b+6ab^2+b^3\right)\)

\(=a^4+6a^3b+6a^2b^2+8ab^3-8a^3b-6a^2b^2-6ab^3-b^4\)

\(=a^4+6a^3b+8ab^3-8a^3b-6ab^3-b^4\)

\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b\right)\left(a-b\right)^3\)

\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)+1=\left(a+b+1\right)^2\)

a) \(24x^2-4xy\)

\(=4x\left(6x-y\right)\)

b) \(5x^3-10x^2+5x-20xy^2\)

\(=5x\left(x^2-10x+5-20y^2\right)\)

nhầm câu b nha 

\(5x^3-10x^2+5x-20xy^2\)

\(=5x\left(x^2-2x+1-\left(2y\right)^2\right)\)

\(=5x\left(\left(x-1\right)^2-\left(2y\right)^2\right)\)

\(=5x\left(x-1-2y\right)\left(x-1+2y\right)\)

\(x^3-25x=0\)

\(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

\(2a^2b\left(a-1\right)+2a\left(1-a\right)\)

\(=2a^2b\left(a-1\right)-2a\left(a-1\right)\)

\(=2a^2b\left(a-1\right)\left(ab-1\right)\)

\(\frac{2m^2+3m+1}{2m^2-m-1}=\frac{2m^2+2m+m+1}{2m^2-2m+m-1}\)

\(=\frac{2m\left(m+1\right)+\left(m+1\right)}{2m\left(m-1\right)+\left(m-1\right)}=\frac{\left(m+1\right)\left(2m+1\right)}{\left(m-1\right)\left(2m+1\right)}\)

\(=\frac{m+1}{m-1}\)