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![](https://rs.olm.vn/images/avt/0.png?1311)
a = x( x3 - 4x2 + 8x - 16)
c, (ab - xy)2 - (bx - ay)2
dựa theo hằng đẳng thức a2 - b2 = ( a + b)(a - b)
=> = [ ( ab - xy) + ( bx - ay)] . [ ( ab - xy) - ( bx - ay) ]
tương tự làm câu d theo câu c
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(x^3+3x^2-3x+1\)
\(=x^3-3x+3x^2-1\)
\(=\left(x-1\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ phân tích thành nhân tử ;
= C2-( a +b )2=( c-a -b ) . ( c+a +b )
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =3x(y-2)-5(y-2)
=(y-2)(3x-5)
b: =(2x-5y)(2x+5y)
c: =x(x^3-8)
=x(x-2)(x^2+2x+4)
a)
\(3x\left(y-2\right)+5\left(2-y\right)\\=3x\left(y-2\right)-5\left(y-2\right)\\ =\left(y-2\right)\left(3x-5\right) \)
b)
\(4x^2-25y^2\\ =\left(2x-5y\right)\left(2x+5y\right)\)
c)
\(x^4-8x\\ =x\left(x^3-8\right)\\ =x\left(x-2\right)\left(x^2+2x+4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
=\(\left(a^2\right)^2-\left(2b\right)^2\)
=\(\left(a^2-2b\right)\left(a^2+2b\right)\)
= \(\left(\left(a-\sqrt{2b}\right)\left(a+\sqrt{2b}\right)\right)\left(a^2+2b\right)\)
c,
=\(4x^4+20x^2+25\)
=\(\left(2x^2\right)^2+2.2x^2.5+5^2\)
=\(\left(2x^2+5\right)^2\)
d,
=\(8x^6-27y^3\)
= \(\left(2x^2\right)^3-\left(3y\right)^3\)
= \(\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
Câu b đề ghi ko rõ lắm
![](https://rs.olm.vn/images/avt/0.png?1311)
2/ (b4 - 4b2 + 4) - 9a2 = (b2 - 2)2 - 9a2 = (b2 - 2 + 3a)(b2 - 2 - 3a)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)
\(=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)\)
\(=\left(9x+1\right)\left(3x-3\right)\)
\(=3\left(9x+1\right)\left(x-1\right)\)
b) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)
\(=\left(8x+11\right)\left(4x+7\right)\)
c) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=-\left[\left(b+c\right)^2-a^2\right]\left(b^2-2bc+c^2-a^2\right)\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left[\left(b-c\right)^2-a^2\right]\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)
d) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-3^2\right]\)
\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b-3\right)\left(a-b+3\right)\)