\(b,x^3-2x^2-4xy^2+x\)

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

\(=x\left[\left(x^2-2x+1\right)-4y^2\right]\)

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

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

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

\(---\)

\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)

Đặt \(y=x^2+7x+10\), thay vào (1) ta được:

\(y\left(y+2\right)-8\)

\(=y^2+2y+1-9\)

\(=\left(y+1\right)^2-3^2\)

\(=\left(y+1-3\right)\left(y+1+3\right)\)

\(=\left(y-2\right)\left(y+4\right)\)

\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)

\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

#Ayumu

a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)

b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)

\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)

\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)

a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)

b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)

=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)

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

Ta có : 5x(x - 2y) + 2(2y - x)²

= 5x(x - 2y) + 2(x - 2y)² (vì (2y - x)² = (x - 2y)² )

= (x - 2y)[5x + 2(x - 2y)]

= (x - 2y)(5x + 2x - 4y)

= (x - 2y)(7x - 4y)

b) 7x(y - 4)² - (4 - y)³ 

= 7x(y - 4)² - (4 - y)²(4 - y)

= 7x(y - 4)² - (y - 4)²(4 - y)

= (y - 4)²(7x - 4 + y)

c) (4x - 8)(x² + 6) - (4x - 8)(x + 7) + 9(8 - 4x)

= (4x - 8)(x² + 6) - (4x - 8)(x + 7) - 9(4x - 8)

= (4x - 8)(x² + 6 - x - 7 - 9)

= 2(x - 4)(x² - x - 10)

\(x^4+4y^4=\left[\left(x^2\right)^2+4x^2y^2+\left(2y^2\right)^2\right]-4x^2y^2=\left(x^2+2y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+8\right)\left(x+12\right)-4x^2\)

\(=\left(x+2\right)\left(x+12\right)\left(x+3\right)\left(x+8\right)-4x^2\)

\(=\left(x^2+14x+24\right)\left(x^2+11x+24\right)-\left(2x\right)^2\)

Đặt \(x^2+11x+24=a\)

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

x⁴+2x³+5x²+4x-12

=x(x³+2x²+5x+4)-12

Có thể chi tiết ra ko bạn, mình cảm ơn.

\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)

\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)

\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)

b) \(3x^2+2x-5=3\left(x-1\right)\left(x+\dfrac{5}{3}\right)\)

c) \(3-2x-x^2=-\left(x-1\right)\left(x+3\right)\)

d) \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e) \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

b: \(3x^2+2x-5\)

\(=3x^2-3x+5x-5\)

\(=\left(x-1\right)\left(3x+5\right)\)

c: \(3-2x-x^2\)

\(=-\left(x^2+2x-3\right)\)

\(=-\left(x+3\right)\left(x-1\right)\)

d: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)

e: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)