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KT
13 tháng 2 2018
Bài 2:
a) Đặt: x - y =a; y - z = b; z - x = c thì a + b + c = 0
C/M: đẳng thức phụ: a3 + b3 + c3 = 3abc
Ta có: \(a+b+c=0\)
\(\Rightarrow\)\(a+b=-c\)
\(\Rightarrow\)\(\left(a+b\right)^3=-c^3\)
\(\Rightarrow\)\(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3=3abc\)
Vậy \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
3 tháng 4 2018
\(a,\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-4\left(x^2y^2-4xy-4\right)\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left[2xy+4\right]^2\)
\(=\left(x^2+y^2-5+2xy+4\right)\left(x^2+y^2-5-2xy-4\right)\)
\(=\left[\left(x^2+y^2+2xy\right)-1\right]\left[\left(x^2+y^2-2xy\right)-9\right]\)
\(=\left[\left(x+y\right)^2-1\right]\left[\left(x-y\right)^2-9\right]\)
\(=\left(x+y-1\right)\left(x+y+1\right)\left(x-y-3\right)\left(x-y+3\right)\)
\(b,x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+8x+4\)
\(=x^2\left(x+1\right)+4\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left[\left(x^2+4\right)\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
\(c,x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left[x^2+2x-3x-6\right]\)
\(=\left(x-5\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(d,125x^3-10x^2+2x-1\)
\(=\left(125x^3-1\right)-\left(10x^2-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
KK
24 tháng 10 2017
a, 4x2 - 4x - 3
=4x2-2x+6x-3
=2x(2x-1)+3(2x-1)
=(2x+3)(2x-1)
b, x3 - x2 - 4
= x3-x2+0x-4
= x3-2x2+x2-2x+2x-4
= (x3-2x2)+(x2-2x)+(2x-4)
= x2(x-2)+x(x-2)+2(x-2)
=(x-2)(x2+x+2)
c, 64x4+y4
=64x4+16x2y2+y4-16x2y2
= (8x2+y2)2-16x2y2
= (8x2+y2-4xy)(8x2+y2+4xy)
TP
0
YF
2
NT
1
22 tháng 8 2017
a)\(\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
b)\(\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
c)\(\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
10 tháng 7 2017
1, \(x^3+8x^2+17x+10=\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)\(=\left(x+1\right)\left(x^2+7x+10\right)=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
2. \(2x^3-3x^2+3x-1=\left(2x^3-x^2\right)-\left(2x^2-x\right)+\left(2x-1\right)\)
\(=x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-x+1\right)\)
3. \(x^4+x^2+1=\left(x^4+1\right)+x^2=\left(x^2+1\right)^2-2x^2+x^2\)\(=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4. \(81x^4+4=\left(9x^2\right)^2+2^2=\left(9x^2+2\right)^2-2.9x^2.2=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)
DT
2
6 tháng 12 2016
a) \(x^3-3x^2+1-3x=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(3x^2-7x-10=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
6 tháng 12 2016
a) \(x^3-3x^2-3x+1=\left(x^3+1\right)-\left(3x^2+3x\right)\)
= \(\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2-x+1-3x\right)\)
= \(\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(3x^2-7x-10=\left(3x^2+3x\right)-\left(10x+10\right)\)
= \(3x\left(x+1\right)-10\left(x+1\right)\)
= \(\left(x+1\right)\left(3x-10\right)\)
1) \(x^2.y^2-1\)
\(=\left(x.y\right)^2-1^2\)
\(=\left(x.y-1\right).\left(x.y+1\right)\)
Chúc bạn học tốt!