1. (a²+b²+ab)²-a²b²-b²c²-c²a²

=a⁴+b⁴+a²b²+2(a²b²+ab³+a³b)-a²b²-b²c²-c²a²

=a⁴+b⁴+2a²b²+2ab³+2a³b-b²c²-c²a²

=(a²+b²)²+2ab(a²+b²)-c²(a²+b²)

=(a²+b²)[(a+b)²-c²]

=(a²+b²)(a+b+c)(a+b-c)

2. a⁴+b⁴+c⁴-2a²b²-2b²c²-2a²c²=(a²-b²-c²)²

3. a(b³-c³)+b(c³-a³)+c(a³-b³)

=ab³-ac³+bc³-ba³+ca³-cb³

=a³(c-b)+b³(a-c)+c³(b-a)

=a³(c-b)-b³(c-a)+c³(b-a)

=a³(c-b)-b³(c-b+b-a)+c³(b-a)

=a³(c-b)-b³(c-b)-b³(b-a)+c³(b-a)

=(c-b)(a-b)(a²+ab+b²)-(b-a)(b-c)(b²+bc+c²)

=(a-b)(c-b)(a²+ab+2b²+bc+c²)

4. a⁶-a⁴+2a³+2a²=a⁴(a+1)(a-1)+2a²(a+1)=(a+1)(a⁵-a⁴+2a²)=a²(a+1)(a³-a²+2)

5. (a+b)³-(a-b)³=(a+b-a+b)[(a+b)²+(a+b)(a-b)+(a-b)²]

=2b(3a²+b²)

6. x³-3x²+3x-1-y³=(x-1)³-y³=(x-1-y)[(x-1)²+(x-1)y+y²]

=(x-y-1)(x²+y²+xy-2x-y+1)

7. x^m+4+x^m+3-x-1=x^m+3(x+1)-(x+1)=(x+1)(x^m+3-1)

(Đúng nhớ like nhá !)

Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi

a) \(6x^2-11xy+3y^2=6x^2-2xy-9xy+3y^2=2x.\left(3x-y\right)-3y.\left(3x-y\right)\)

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

b) PP: dùng hệ số bất định

ta có: x^4 -3x^3+6x^2-5x+3=(x^2+ax-1)(x^2 +bx-3)  (*)

                                           =x^4 +bx^3-3x^2+ax^3 +(a+b)x^2 -3ax  -x^2-bx+3

                                           =x^4 +(b+a)x^3 +(a+b-3-1)x^2 -(3a+b)x +3

=> a+b=-3

    a+b-4=6          

   3a+b=5

<=> a=7/2 ;b=13/2  thay vào (*) ta đc: x^4 -3x^3+6x^2-5x+3=(x^2+\(\frac{7}{2}\).x -1)(x^2 +\(\frac{13}{2}\).x -3)

Hay x^4 -3x^3+6x^2-5x+3= \(\frac{1}{4}.\left(2x^2+7x-2\right)\left(2x^2+13-6\right)\)

het thoirui pan oi

a) A=(4-5x)²-(3+5x)²=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

B=(3x-1)(1+3x)-(3x+1)²=9x²-1-(3x+1)²=9x²-1-(9x²+6x+1)=9x²-1-9x²-6x-1=-6x-2=-2(3x+1)

1.

\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)

Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)

Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)

Từ đó ta được đpcm

2.

\(a,Sửa:a^6+a^4+a^2b^2+b^4-b^6\\ =\left(a^6-b^6\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+b^4+a^2b^2\right)\\ =\left(a^2-b^2+1\right)\left(a^4+a^2b^2+b^4\right)\\ =\left[\left(a^2+b^2\right)^2-a^2b^2\right]\left(a^2-b^2+1\right)\\ =\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^2-b^2+1\right)\\ b,=\left(a^3+b^3\right)-1+3ab\\ =\left(a+b\right)^3-3ab\left(a+b\right)-1+3ab\\ =\left(a+b-1\right)\left(a^2+2ab+b^2+a+b+1\right)-3ab\left(a+b-1\right)\\ =\left(a+b-1\right)\left(a^2+b^2+1+a+b-ab\right)\)

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

a/ (x - 1)(x - √3 + 2)(x + √3 + 2)

a ) \(x^3+3x^2-3x+1\)

    \(=x^3-3x+3x^2-1\)

     \(=\left(x-1\right)^3\)

c/ (x + 1)(x + \(\frac{\sqrt{21}-5}{2}\))(x + \(\frac{-\sqrt{21}-5}{2}\))

a + b = c => (a + b)² = c² <=> a²+ b² + 2ab = c² 
=> c^4 = (a² + b² + 2ab)² 
=> c^4 = a^4 + b^4 + 6a²b² + 4a^3.b + 4a.b^3 

vậy: a^4 + b^4 + c^4 = 2a^4 + 2b^4 + 6a²b² + 4a^3.b + 4a.b^3 
= 2a^4 + 2a²b² + 4a^3.b + 2b^4 + 2a²b² + 4a.b^3 + 2a²b² 
= 2a²(a² + b² + 2ab) + 2b²(b² + a² + 2ab) + 2a²b² 
= 2a²(a + b)² + 2b²(a + b)² + 2a²b² 
= 2a²b² + 2(a + b)²(a² + b²) 
= 2a²b² + 2c²(a² +b²) 
= 2a²b² + 2b²c² + 2c²a² (đpcm)