11, \(5\left(x-y\right)+y^2-xy=5\left(x-y\right)+y\left(y-x\right)=\left(5-y\right)\left(x-y\right)\)

12, \(14a^2b^3-35ab^2+7a^2b^2=7ab^2\left(2ab-5+a\right)\)

13, \(6a^2b^3-30ab^2+12a^2b=6ab\left(ab^2-5b+2a\right)\)

6, \(x^2-1+2xy+y^2=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)

7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)

8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)

9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)

10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)

1 x mũ 2 + 4xy + 4y mũ 2 = x^2 + 4xy + 4y^2 =(2y+x)^2

2,       4x mũ 2 - 36y mũ 2 =4x^2 -36y^2 = -4 (3y-x) (3y+x)

6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)

8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2  = -(y-x^2-x) (y+x^2+x)

10, 4x mũ 2 ( x + y ) -x - y  = (2x-1) (2x+1) (y+x)

Ta có x + y = a + b 

=> (x + y)² = (a + b)² 

=> x² + y² + 2xy = a² + b² + 2ab 

=> xy = ab

Lại có x + y = a + b

=> (x  + y)³ = (a + b)³ 

=> x³ + 3x²y + 3xy² + y³ = a³ + 3a²b + 3ab² + b³ 

=> x³ + y³ + 3xy(x + y) = a³ + b³ + 3ab(a + b)

=> x³ + y³ = a³ + b³ (vì x + y = a + b ; xy = ab)

\(\left(5n-7\right)^2-9\)

\(=\left(5n-7\right)^2-3^2\)

\(=\left(5n-7-3\right)\left(5n-7+3\right)\)

\(=\left(5n-10\right)\left(5n-4\right)\)

\(5\left(n-2\right)\left(5n-4\right)⋮5\)với mọi số nguyên n \(\left(đpcm\right)\)

a)\(=5c\left(a^2-9b^2\right)+5c^2.\left(3b-a\right)\))\

\(=5c\left(a-3b\right)\left(a+3b\right)-5c^2.\left(a-3b\right)\)

= \(5c.\left(a-3b\right)\left(a+3b-c\right)\)

b)\(=xy.\left(3x-y\right)-xyz.\left(3x-y\right)\)

\(=xy.\left(3x-y\right)\left(1-z\right)\)

c) \(=a^2m^2+2ambp+b^2p^2-a^2p^2-2ambp-b^2m^2\)

\(=a^2m^2-a^2p^2+b^2p^2-b^2m^2\)

\(=a^2.\left(m^2-p^2\right)+b^2\left(p^2-m^2\right)\)

\(=a^2.\left(m^2-p^2\right)-b^2\left(m^2-p^2\right)\)

\(=\left(a^2-b^2\right)\left(m^2-p^2\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(m-p\right)\left(m+p\right)\)

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

\(=\frac{x^6}{y^9}-3.\frac{x^4}{y^6}.\frac{y^3}{z^2}+3\frac{x^2}{y^3}.\frac{y^9}{z^4}-\frac{y^9}{z^6}=\frac{x^6}{y^9}-\frac{3x^4}{y^3.z^2}+\frac{3x^2y^6}{z^4}-\frac{y^9}{z^6}\)

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

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

\(=\frac{x^6}{y^9}-3.\frac{x^4}{y^6}.\frac{y^3}{z^2}+3.\frac{x^2}{y^3}.\frac{y^6}{z^4}-\frac{y^9}{z^6}\)

\(=\frac{x^6}{y^9}-\frac{3x^4}{y^3z^2}+\frac{3x^2y^3}{z^4}-\frac{y^9}{z^6}\)

\(=\frac{x^6z^6-3x^4y^6z^4+3x^2y^{12}z^2-y^{18}}{y^9z^6}\)

\(a)\)

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

\(\Leftrightarrow\left(9-6x+x^2\right)-\left(x^2-4x\right)=2x-5\)

\(\Leftrightarrow9-2x=2x-5\)

\(\Leftrightarrow4x=14\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(b)\)

\(x^2-2x+1=25x^2\)

\(\Leftrightarrow x^2-2x+1-25x^2=0\)

\(\Leftrightarrow-24x^2-2x+1=0\)

\(\Leftrightarrow-\left(24x^2+2x-1\right)=0\)

\(\Leftrightarrow24x^2+2x-1=0\)

\(\Leftrightarrow x=\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-1}{4}\end{cases}}\)

\(c)\)

\(4x^2-4x=24\)

\(\Leftrightarrow x^2-x=6\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow x=\orbr{\begin{cases}x=\left(-2\right)\\x=3\end{cases}}\)

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

\(< =>\left(9-6x+x^2\right)-\left(x^2-4x\right)=2x-5\)

\(< =>9-6x+x^2-x^2+4x=2x-5\)

\(< =>-6x+4x-2x=-5-9\)

\(< =>-4x=-14\)

\(< =>x=\frac{7}{2}\)

b)\(x^2-2x+1=25x^2\)

\(< =>\left(x-1\right)^2-25x^2=0\)

\(< =>\left(x-1\right)^2-\left(5x\right)^2=0\)

\(< =>\left(x-1-5x\right)\left(x-1+5x\right)=0\)

\(< =>\orbr{\begin{cases}-4x-1=0\\6x-1=0\end{cases}}\)

\(< =>\orbr{\begin{cases}-4x=1\\6x=1\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{6}\end{cases}}\)

c)\(4x^2-4x=24\)

câu này mk ko biết làm @@ sorry nha :))