Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

Bài 1;

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

2) \(6x^2+12xy+6y^2=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\)

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

4) \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

3) \(a^3b-ab^3+a^2+2ab+b^2=ab\left(a^2-b^2\right)+\left(a+b\right)^2=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)

4) \(2x^3-2xy^2-8x^2+8xy=2x\left(x^2-y^2-4x+4y\right)=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]=2x\left(x-y\right)\left(x+y-4\right)\)

1.  \(xy\left(a^2+2b^2\right)-ab\left(2x^2+y^2\right)\)

\(=xya^2+2xyb^2-2abx^2-aby^2\)

\(=xya^2-aby^2-2abx^2+2xyb^2\)

\(=ay\left(ax-by\right)-2bx\left(ax-by\right)\)

\(=\left(ay-2bx\right)\left(ax-by\right)\)

2. \(xy\left(a^2+2b^2\right)+ab\left(2x^2+y^2\right)\)

\(=xya^2+2xyb^2+2abx^2+aby^2\)

\(=xya^2+aby^2+2abx^2+2xyb^2\)

\(=ay\left(ax+by\right)+2bx\left(ax+by\right)\)

\(=\left(ay+2bx\right)\left(ax+by\right)\)

a) 16 + 2x³y³

= 2( 8 + x³y³)

= 2[ 2³ + (xy)³]

= 2 (2+xy)(4 - 2xy + x²y²)

b) 100a² - (a² + 25)²

= (10a)² - (a² +25)²

= (10a - a² - 25)(10a + a² +25)

= -(a² - 2a.5 + 5²)(a² + 2a.5 + 5²)

=-(a-5)² (a+5)²

a) \(16+2x^3y^3\)

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

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

b) \(100a^2-\left(a^2+25\right)^2\)

\(=\left(10a-a^2-25\right)\left(10a+a^2+25\right)\)

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

= (4y)^2 + 12ax - (3x)^2 - (2a)^2

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

= (4y-3x-2a)(4y-3x+2a) 

a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)

\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b² thành 27b³ vì mình nghĩ nhầm đề)

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

\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)

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

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

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

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

a)  36-4a²+20ab-25b²

= 6^2 - (4a^2 - 20xb + 25b^2)

= 6^2 - (2a - 5b)^2

= [6 - (2a - 5b)] [6 + (2a - 5b)]

= (6 - 2a + 5b) (6 + 2a -5b)

a) \(6^2-\left(\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right)\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

b) \(=\left(a+1\right)^3-\left(3b\right)^3\)

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

\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)

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

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

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

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

\(\left(a+4\right)^2-16a^2\)

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

\(=\left(a+4+4a\right)\left(a+4-4a\right)\)

\(=\left(5a+4\right)\left(4-3a\right)\)

\(x^3=4x\)

\(\Leftrightarrow x^3-4x=0\)

\(\Leftrightarrow x\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{4}=\pm2\end{cases}}\)

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)

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

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

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

\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)

 TL:

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

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

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

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

=(x+2)

mk lm tp nè

b)(x+1)\(^2\)+(x+3)(x-3)=(x+1)\(^2\)-(x+3)(x+3)

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

=(x+1+x+3)(x+1-x+3)

=(2x+4)4

a) 27x^3 –27x^2 +18x –4

= 27x^3 –9x^2–18x^2+6x + 12x –4

= 9x^2 (3x–1) – 6x (3x–1) +4(3x–1)

= (3x-1) (9x^2–6x+4)

b)2x^3–2x^2+5x+3

= 2x^3+x^2–2x^2–x+6+3

= x^2(2x+1)-x^2(2x+1)+3(2x+1)

= (2x+1) 3

c) 2x^4 + 5x^3+13x^2+25x+15 
=2x^3(x+1)+3x^2(x+1)+10x(x+1)+15(x+1) 
=(x+1)(x^2(2x+3)+5(2x+3)) 
=(x+1)(2x+3)(x^2+5)