\(x^2-y^2+5x-5y\)

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

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

\(---\)

\(x^2-16y^2+4x+4\)

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

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

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

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

\(---\)

\(3x^2+6xy+3y^2-12\)

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

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

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

\(---\)

\(4x^3+4x^2+x\)

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

\(=x\left(2x+1\right)^2\)

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

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

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)

mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk

\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

câu thứ 2 bạn ktra lại đề

\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)

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

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

a)  \(x^5-7x^4-x^3+43x^2-36\)

\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-6\right)\)

c)  \(x^4+2x^3-15x^2-18x+64\)

\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)

\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)

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

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

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(b,x^3-3x^2-4x+12\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-4\right)\)

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

\(c,3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)

\(\text{d) 2x}^4- 7x^3 - 2x^2 + 13x + 6\)
\(\text{= (2x^4 + 2x^3) - (9x^3 + 9x^2) + (7x^2 + 7x) + (6x + 6)}\)
\(\text{= 2x^3(x + 1) - 9x^2(x + 1) + 7x(x + 1) + 6(x + 1)}\)
\(\text{= (x + 1)(2x^3 - 9x^2 + 7x + 6)}\)
\(\text{= (x + 1)(2x + 1)(x - 3)(x - 2)}\)

phần b,c thay ''<=>'' là ''='' nhé ! Mình nhầm!

\(a,=x\left(x+5\right)\\ b,=\left(y-1\right)^2-4x^2=\left(y-1-2x\right)\left(y-1+2x\right)\\ c,\text{Không phân tích đc}\)

a) \(x^{2}+5x=x(x+5)\)

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

c)\(x^{2}-8x+16=(x-4)^{2}\)