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\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
a)(x+y)2-(x-y)2
=(x+y-x+y)(x+y+x-y)
=2y.2x=4xy
b)(3x+1)2-(x+1)2
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
=4x(2x+1)
c) x3+y3+z3-3xyz
= (x+y)3- 3xy(x+y) +z3-3xyz
=(x+y+z)( x2+2xy+y2-xz-yz+z2)-3xy(x+y+z)
=(x+y+z)(x2+y2+z2-xy-xz-yz)
Phân tích đa thức sau thành nhân tử :
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b) \(x^3+y^3+z^3-3xyz\)
\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)
\(=\left(y-1\right)\left[\left(\frac{2}{5}x-\frac{2}{5}y\right)\right]\)
\(=\left(y-1\right)\frac{2}{5}\left(x-y\right)\)
1) \(VT=x^3+x^2y-x^2y-xy^2+xy^2+y^3=x^3+y^3=VP\)
2) \(VP=x^2+xy-xy-y^2=x^2-y^2=VT\)
3) \(VP=x^2+2\cdot x\cdot1+1=x^2+2x+1=VT\)
4) \(VP=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3=VT\)
1, \(\left(x^2-xy+y^2\right)\left(x+y\right)=x^3+y^3\\ x^3+x^2y-x^2y-xy^2+xy^2+y^3=x^3+y^3\\ x^3+y^3=x^3+y^3\left(đúng\right)\)Vậy ta được đpcm
2, \(x^2-y^2=\left(x-y\right)\left(x+y\right)\\ x^2-y^2=x^2+xy-xy-y^2\\ x^2-y^2=x^2-y^2\left(đúng\right)\)Vậy ta được đpcm
3, \(x^2+2x+1=\left(x+1\right)^2\\ x^2+2x+1=\left(x+1\right)\left(x+1\right)\\ x^2+2x+1=x^2+x+x+1\\ x^2+2x+1=x^2+2x+1\left(đúng\right)\)Vậy ta được đpcm
4, \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\\ x^3-y^3=x^3+x^2y+xy^2-x^2y-xy^2-y^3\\ x^3-y^3=x^3-y^3\left(đúng\right)\)Vậy ta được đpcm
Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...
a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)
\(=4\left[\left(x^2-2x+1\right)-\left(a^2-2ay+y^2\right)\right]\)
\(=4\left[\left(x-1\right)^2-\left(a-y\right)^2\right]\)
\(=4\left(x-1-a+y\right)\left(x-1+a-y\right)\)
b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
c)\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x^2-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(5x+5\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x-1\right)\left(x+3\right)^2\)
d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)
\(=\left(a+1\right)\left(a^4+a^2+1\right)\)
\(=\left(a+1\right)\left(a^4+2a^2+1-a^2\right)\)
\(=\left(a+1\right)\left[\left(a^2+1\right)^2-a^2\right]\)
\(=\left(a+1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)\)
e)\(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-1-y\right)\left(x^2-2x+1+xy-y+y^2\right)\)
f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)
\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y\right)\left(5x-3y\right)\)
g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)
\(=3\left(a-b+c\right)\left(x+6y\right)^2\)
a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)
\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)
\(=\left(2x-2\right)^2-\left(2y-2a\right)^2=\left(2x-2+2y-2a\right)\left(2x-2-2y+2a\right)\)
b/ \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)=\left(x+y-1\right)\left(x^2+y^2+2xy+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)
Giải giúp bạn 2 bài tiêu biểu thôi nha
a)\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)
\(\left(x-y\right)^3-\left(x+y\right)^3\\ =\left(x-y-x-y\right)\left(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2\right)\\ =-2y\left(3x^2+y^2\right)\)
\(\left(x+1\right)^3+\left(x-1\right)^3\\ =\left(x+1+x-1\right)\left(x^2+2x+1-x^2+1+x^2-2x+1\right)\\ =2x\left(x^2+3\right)\)
\(\left(x-1\right)^2-\left(x+1\right)^2\\ =\left(x-1-x-1\right)\left(x-1+x+1\right)\\ =-2.2x=-4x\)
a: =(2x)^3+3*(2x)^2*3y+3*2x*(3y)^2+(3y)^3
=(2x+3y)^3
b: (x-y)^3-(x+y)^3
=(x-y-x-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]
=-2y*[x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2]
=-2y(3x^2+y^2)
c: (x+1)^3+(x-1)^3
=(x+1+x-1)[(x+1)^2-(x+1)(x-1)+(x-1)^2]
=2x*[x^2+2x+1-x^2+1+x^2-2x+1]
=2x(x^2+3)
d: =(x-1-x-1)(x-1+x+1)
=2x*(-2)=-4x