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a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x-6y-1\right)\)
b) \(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c) \(=2\left(x-y\right)^2-18\)
\(=2\left[\left(x-y\right)^2-3^2\right]\)
\(=2\left(x-y+3\right)\left(x-y-3\right)\)
a: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: \(x^3-8x^2+16x\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: Ta có: \(16x-8x^2+x^3\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: Ta có: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\cdot\left[\left(x-y\right)^2-9\right]\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: Ta có: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
e: Ta có: \(x^4-x^2-30\)
\(=x^4-6x^2+5x^2-30\)
\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)
\(=\left(x^2-6\right)\left(x^2+5\right)\)
f: Ta có: \(x^2-xy-2y^2\)
\(=x^2-2xy+xy-2y^2\)
\(=x\left(x-2y\right)+y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+y\right)\)
g: Ta có: \(x^4-13x^2y^2+4y^4\)
\(=x^4-4x^2y^2+4y^4-9x^2y^2\)
\(=\left(x^2-2y^2\right)^2-\left(3xy\right)^2\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-3xy+2y^2\right)\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-xy-2xy+2y^2\right)\)
\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)
\(=\left(x-y\right)\left(x-2y\right)\left(x^2-3xy-2y^2\right)\)
h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)^2-3\left(x^2-2x\right)+\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x^2-2x+1\right)\)
\(=\left(x-3\right)\left(x+1\right)\cdot\left(x-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x^2-7x+12\right).\left(x^2-15x+56\right)-60\)
\(=\left(x-3\right)\left(x-4\right)\left(x-7\right)\left(x-8\right)-60\)
b) \(x^4+2000x^2+1999x+2000\)
\(=\left(x^2-x+2000\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x-1\right)^2+1999\left(x^2+x+1\right)+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$
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$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$
$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.
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$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$
$=3(x-2)(x^2+2x+4)-240x$
$=3(x^3-2^3)-240x=3x^3-240x-24$
$=3(x^3-80x-8)$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
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x4+4 = (x2)2+22 = x4 + 2.x2.2 + 4 – 4x2
= (x2 + 2)2 – (2x)2 = (x2-2x+2)(x2+2x+2)
Ta có: \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x4 + 4
= (x2)2 + 22
= x4 + 2.x2.2 + 4 – 4x2
(Thêm bớt 2.x2.2 để có HĐT (1))
= (x2 + 2)2 – (2x)2
(Xuất hiện HĐT (3))
= (x2 + 2 – 2x)(x2 + 2 + 2x)
https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-x-4-16-thanh-nhan-tu-faq324398.html
\(x^4+16x-12\)
\(=x^4+2x^3-2x^2-2x^3-4x^2+4x+6x^2+12x-12\)
\(=x^2.\left(x^2+2x-2\right)-2x.\left(x^2+2x-2\right)+6.\left(x^2+2x-2\right)\)
\(=\left(x^2+2x-2\right)\left(x^2-2x+6\right)\)