bài 1 :phân tích các đa thức sau thành nhân tử
a)x6- y6
b)6x-9-x
c)x2+4y2+4xy
d)(x+y)2-(x-y)2=[(x+y)-(x-y)]
e)(4x+3)2-(2x-1)2
f)x^3+y^3+z^3-3xy
g)x^3-2xy+y^2-z^2
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a,x^2-x-y^2-y
=x^2-y^2-(x+y)
=(x-y).(x+y)-(x+y)
=(x+y).(x-y-1)
b, x^2-2xy+y^2-z^2
=(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
c,5x-5y+ax-ay( đề bài ở đây phải là -ay ms tính đc)
=(5x-5y)+(ax-ay)
=5(x-y)+a(x-y)
=(x-y).(5+a)
d,a^3-a^2.x-ay+xy
=(a^3-a^2x)-(ay-xy)
=a^2(a-x)-y(a-x)
=(a-x)(a^2-y)
e,4x^2-y^2+4x+1
={(2x)^2+4x+1}-y^2
=(2x+1)^2-y^2
=(2x+1+y^2)(2x+1-y^2)
f,x^3-x+y^3-y
=(x^3+y^3)-(x+y)
=(x+y)(x^2-xy+y^2)-(x+y)
=(x+y)(x^2-xy+y^2-1)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
a)\(=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
b)\(=\left(x+9\right)^2-\left(6x\right)^2=\left(x+9-6x\right)\left(x+9+6x\right)=\left(-5x+9\right)\left(7x+9\right)\)
c)\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)=\left(x-y\right)^2-\left(z-t\right)^2\\ =\left(x-y+z-t\right)\left(x-y-z+t\right)\)
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
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: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
a) (x - 1)(x + l)(x - 2)(x - 4). b) (x - 2)( x 2 + 4).
c) 2y(3 x 2 + y 2 ). d) 2(x + y + z) ( a - b ) 2 .
a. \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)
\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-3\right)^2-1\right]\left(x^2-1\right)\)
\(=\left(x-3+1\right)\left(x-3-1\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)
b. \(x^3-2x^2+4x-8\)
\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)
\(=x\left(x^2+4\right)-2\left(x^2+4\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
c. \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)
\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)
\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)
\(=2\left(a-b\right)^2\left(x+y+z\right)\)
a: \(2x^2+3xy-14y^2\)
\(=2x^2+7xy-4xy-14y^2\)
\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)
\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)
\(=\left(2x+7y\right)\left(x-2y\right)\)
b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)
\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)
\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)
\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)
\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)
c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)
\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)
\(=\left(7x-5\right)\left(-2x-2\right)\)
\(=-2\left(x+1\right)\left(7x-5\right)\)
d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)
\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)
\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)
\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)
\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)
\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)
\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
a) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
b) sửa đề nhé!
\(6x-9-x^2=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)