đk:   \(x\ne0\);  \(x\ne\pm3y\)

\(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

\(=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

\(=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{\left(x+3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\frac{x+3y}{x\left(x-3y\right)}\)

\(\frac{x^2+25y^2-10xy}{x^2-25y^2}\)

\(\frac{\left(x-5y\right)^2}{\left(x-5y\right)\cdot\left(x+5y\right)}\)

\(\frac{x-5y}{x+5y}\)

đk:  \(x\ne\pm5y\)

\(\frac{x^2+25y^2}{x^2-25y^2}-\frac{10xy}{x^2-25y^2}\)

\(=\frac{x^2+25y^2-10xy}{\left(x-5y\right)\left(x+5y\right)}\)

\(=\frac{\left(x-5y\right)^2}{\left(x-5y\right)\left(x+5y\right)}\)

\(=\frac{x-5y}{x+5y}\)

\(x^2-3x-4x+12+x^2+10x+25\)

\(2x^2-7x+10x+27\)

\(2x^2+3x+27\)

\(\left(x-4\right)\left(x-3\right)+\left(x+5\right)^2\)

\(=x^2-7x+12+x^2+10x+25\)

\(=2x^2+3x+27\)

\(\left(a^2+9\right)^2-36a^2\)

\(=\left(a^2+9+6a\right)\left(a^2+9-6a\right)\)

\(=\left(a+3\right)^2.\left(a-3\right)^2\)

p/s:chúc bạn hk tốt

\(\left(a^2+9\right)^2-36a^2\)

\(=\left(a^2+9+36a^2\right)\cdot\left(a^2+9-36a^2\right)\)

\(=\left(37a^2+9\right)\cdot\left(-35a^2+9\right)\)

\(\left(a^2+4b^2\right)^2-16a^2\)

\(=\left(a^2+4b^2-4a\right)\left(a^2+4b^2+4a\right)\)

p/s: chúc bạn học tốt

(a^2+4b^2)^2-16a^2

=(a^2+4b^2-4a)(a^2+4b^2+4a)

\((4x-y)(a+b)(4x-y)(c-1)\)

\(=\left(4x-y\right)\left(4x-y\right)=\left(4x-y\right)^{1+1}=\left(4y-2\right)^2\)

\(=\left(a+b\right)\left(4x-y\right)^2\left(c-1\right)\)

(4x-y)(a+b)(4x-y)(c-1)

= ( 4x - y ) ( 4x - y ) = ( 4x - y ) ^{1 + 1} = ( 4y - 2 ) ²

= (a + b ) ( 4x - y )²  ( c - 1 )

đa thức này đã phân tích thành nhân tử rồi bạn ạ