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

1. 5x² - 4x

= x(5x - 4)

2. 8x²(x - 3y) - 12x(x - 3y)

= (8x² - 12x)(x - 3y)

= 4x(2x - 3)(x - 3y)

3. 3(x - y) - 5x(y - x)

= 3(x - y) + 5x(x - y)

= (3 + 5x)(x - y)

a: =(16x+20)^2-(10x+10)^2

=(16x+20-10x-10)(16x+20+10x+10)

=(26x+30)(6x+10)

=4(13x+15)(3x+5)

b: =(x-y+4-2x-3y+1)(x-y+4+2x+3y-1)

=(-x-4y+5)(3x+2y+3)

c: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

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

=2(x^2+1)*4x

=8x(x^2+1)

$16(4x+5)^2-25(2x+2)^2$

$=[4(4x+5)]^2-[5(2x+2)]^2$

$=[4(4x+5)-5(2x+2)][4(4x+5)+5(2x+2)]$

$=(16x+20-10x-10)(16x+20+10x+10)$

$=(6x+10)(26x+30)$

$=4(3x+5)(13x+15)$

$(x-y+4)^2-(2x+3y-1)^2$

$=[(x-y+4)-(2x+3y-1)][(x-y+4)+(2x+3y-1)]$

$=(-x-4y+5)(3x+2y+3)$

Vậy:

$=-(x+4y-5)(3x+2y+3)$

$(x+1)^4-(x-1)^4$

$=[(x+1)^2]^2-[(x-1)^2]^2$

$=[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]$

$=[x^2+2x+1-(x^2-2x+1)][x^2+2x+1+x^2-2x+1]$

$=(4x)(2x^2+2)$

$=8x(x^2+1)$

d)

Câu này không phân tích được thành nhân tử với hệ số nguyên/rational theo các hằng đẳng thức thông thường: $x^6-y^5$

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

x⁴ - 8x = x.(x³ - 8) = x. (x³ - 2³) = x.(x - 2).(x² + 2x +4)

Ai trả lời mình đi cho đỡ quê 😞💔

a: \(=\left(x+2-y\right)\left(x+2+y\right)\)

c: \(=\left(x-y\right)^2\)

a)\(6x^2-9xy\)

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

b)\(x^2-y^2-3x+3y\)

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

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

c)\(x^4-8x^2-9\)

\(=x^4+x^2-9x^2-9\)

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

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

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

d)\(x^4-4\left(x^2+5\right)-25\)

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

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

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

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

a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)

b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)

\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)

\(=\left(2x+y\right)\left(4x^2-8xy+y^2\right)\)

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

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

\(a,=\left(x+y\right)\left(x+3\right)\\ b,=\left(x+2\right)\left(x+3\right)\)

\(\left(3x+1\right)^2-4\left(x-2\right)^2=9x^2+6x+1-4\left(x^2-4x+4\right)=9x^2+6x+1-4x^2+16x-16=5x^2+22x-15=\)

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

\(9\left(2x+3\right)^2-4\left(x+1\right)^2=9\left(4x^2+12x+9\right)-4\left(x^2+2x+1\right)=36x^2+108x+81-4x^2-8x-4=32x^2+100x+77\)

\(\left(8x+11\right)\left(4x+7\right)\)