phân tích đa thức thành nhân tử
a)a^3-b^3+c^3+3abc
b)a^3 -b^3-c^3-3abc
c)(x-y)^3+(y-z)^3+(z-x)^3
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Khách
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3 tháng 9 2023
a: =(x+y)^3+z^3-3xy(x+y)-3xyz
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2
=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c
=b^2(c-a)+b(c^2-a^2)+ac(c-a)
=(c-a)(b^2+ac)+b(c-a)(c+a)
=(c-a)(b^2+ac+bc+ba)
=(c-a)[b^2+bc+ac+ab]
=(c-a)[b(b+c)+a(b+c)]
=(c-a)(b+c)(b+a)
NL
0
18 tháng 10 2021
\(a,\Rightarrow\left(x-3-5+2x\right)\left(x-3+5-2x\right)=0\\ \Rightarrow\left(3x-8\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\\ b,=\left(x+y\right)^2-\left(x-2y\right)^2\\ =\left(x+y-x+2y\right)\left(x+y+x-2y\right)=3y\left(2x-y\right)\\ c,=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\\ d,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
NB
1
T
29 tháng 8 2019
#)Giải :
a) \(x+y+z=0\Leftrightarrow x+y=-z\Leftrightarrow\left(x+y\right)^3=\left(-z\right)^3\Leftrightarrow x^3+3x^2y+3xy^2+y^3=\left(-z\right)^3\)
\(\Leftrightarrow x^3+y^3+z^3=-3x^2y-3xy^2\Leftrightarrow x^3+y^3+z^3=-3xy\left(-z\right)\) hay 3xyz (đpcm)
b) \(x=\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(\Leftrightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) (Áp dụng hằng đẳng thức)
\(\Leftrightarrow x=\left[\left(b-c\right)^3+\left(c-a\right)^3\right]+\left(a-b\right)^3\)
\(=\left[\left(b-a\right)^3+\left(c-a\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[\left(b-c\right)+\left(c-a\right)\right]+\left(a-b\right)^3\)
\(=\left(b-a\right)^3-3\left(b-c\right)\left(c-a\right)\left(b-a\right)+\left(a-b\right)^3\)
\(=\left[-\left(a-b\right)^3\right]-3\left(b-c\right)\left(c-a\right)\left[-\left(a-b\right)\right]+\left(a-b\right)^3\)
\(=-\left(a-b\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)+\left(a-b\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
5 tháng 10 2021
\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)
\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
HT
0
ta có :
\(a^3+c^3=\left(a+c\right)^3-3ac\left(a+c\right)\)
nên \(a^3+c^3-b^3+3abc=\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)\)
\(=\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2-3ac\right]=\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)\)
b. tương tự ta có :
\(a^3-b^3-c^3-3abc=a^3-\left(b+c\right)^3+3bc\left(b+c-a\right)\)
\(=\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2-3bc\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
c. ta có : \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=\left(x-z+z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+3\left(x-z\right)\left(z-y\right)\left(x-y\right)+\left(z-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=3\left(x-z\right)\left(z-y\right)\left(x-y\right)\)