Ta có:\(b^4+4a^4=b^4+4a^2b^2+4a^4-4a^2b^2\)

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

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

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

a2 – b2 – 4a + 4

= a2 – 4a + 4 – b2

= (a – 2)2 – b2

= (a – 2 + b)(a – 2 – b)

= (a + b – 2)(a – b – 2)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4.\)

\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4.\)

\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4.\)

\(=\left(x+5ax+4a^2+a^2\right)^2.\)

\(=\left(x+5ax+5a^2\right)^2.\)

\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

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

\(=\)\(\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

\(=\)\(\left[\left(x^2+5ax+5a^2\right)-a^2\right].\left[\left(x^2+5ax+5a^2\right)-a^2\right]+a^4\)

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

\(=\)\(\left(x^2+5ax+5a^2\right)^2\)

Chúc bạn học tốt ~ 

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

3n(m - 3) + 5m(m - 3)

= (3n + 5m)(m - 3)

2a(x - y) - (y - x)

= (x - y)(2a + 1)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

4a²b² + 36a²b³ + 6ab⁴

= 2ab²(2a + 18ab + 3b²)

4a²b³ - 6a³b²

= 2a²b²(2b - 3a)

a) x⁴ - y⁴

= (x²)² - (y²)²

= (x² - y²)(x² + y²)

= (x - y)(x + y)(x² + y²)

b) 1 - 8x³y⁶

= 1³ - (2xy²)³

= (1 - 2xy²)(1 + 2xy² + 4x²y⁴)

\(a^3+4a^2-7a-10\)

\(=\left(a^3+5a^2\right)-\left(a^2+5a\right)-\left(2a+10\right)\)

\(=a^2\left(a+5\right)-a\left(a+5\right)-2\left(a+5\right)\)

\(=\left(a^2-a-2\right)\left(a+5\right)\)

\(=\left(a^2-2a+a-2\right)\left(a+5\right)\)

\(=\left[a\left(a-2\right)+\left(a-2\right)\right]\left(a+5\right)\)

\(=\left(a+1\right)\left(a-2\right)\left(a+5\right)\)