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1)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(b2c-b2a)+(c2a-a2c)
=b.(a2-c2)-b2.(a-c)-ac.(a-c)
=b.(a-c)(a+c)-b2(a-c)-ac(a-c)
=(a-c)(ab+bc-b2-ac)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
Bài 3a)
\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
mà \(a+b=-c\Rightarrow a^3+b^3+c^3=3abc\)
\(1,\)\(3x-3y-x^2+2xy-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(3x-3y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(3-x+y\right)\)
\(2,\)\(49\left(x-4\right)^2-9\left(x+2\right)^2\)
\(=\left[7\left(x-4\right)\right]^2-\left[3\left(x-2\right)\right]^2\)
\(=\left(7x-28-3x+6\right)\left(7x-28+3x-6\right)\)
\(=\left(4x-22\right)\left(10x+34\right)\)
\(3,\)\(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2+5\right)\left(x-1\right)\left(x+1\right)\)
- x2.(x3-x2+x-1)
- x.( x3-3x2-1)+3
- x.(x2-xy-y2)
Tìm x:
x3-16x = 0
=> x.(x2-16) = 0
=> x = 0 hay x2-16 = 0
=> x = 0 hay x2 = 0+16
=> x = 0 hay x2 = 16
=> x = 0 hay x = 4 hay x = -4
a) \(x^3+3.2x^2y+3.2^2.x.y^2+\left(2y\right)^3=\left(x+2y\right)^3\)
b) áp dụng HDT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
c) cũng áp dụng hdt :\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2=\left[3\left(x+5\right)-x+7\right]\left[3\left(x+5\right)+x-7\right]\)\(=\left(3x+15-x+7\right)\left(2x+15+x-7\right)=\left(2x+22\right)\left(3x+8\right)=2\left(x+11\right)\left(3x+8\right)\)
d) áp dụng típ \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)=\left(x-9y\right)\left(9x-y\right)\)
e)Áp dụng típ Hdt như trên
\(\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)=\left(4y-34\right)\left(11y-22\right)\)
\(=2\left(2y-17\right).11\left(y-2\right)=22\left(2y-17\right)\left(y-2\right)\)
Bạn 1 cái t i c k nha thật sự rất cảm ơn
3x^2+2x-1
=3x^2+3x-x-1
=3x(x+1)-(x+1)
=(x+1)(3x-1)
x^3+6x^2+11x+6
=x^3+5x^2+6x+x^2+5x+6
=x(x^2+5x+6)+(x^2+5x+6)
=(x+1)(x^2+5x+6)
=(x+1)(x^2+3x+2x+6)
=(x+1)(x+2)(x+3)
x^4+2x^2-3
=x^4-x^2+3x^2-3
=x^2(x^2-1)+3(x^2-1)
=(x^2-1)(x^2+3)
=(x+1)(x-1)(x^2+3)
ab+ac+b^2+2bc+c^2
=a(b+c)+(b+c)^2
=(b+c)(a+b+c)
a^3-b^3+c^3+3abc
=(a-b)^3+3ab(a-b)+c^3+3abc
=(a-b+c)^3-3(a-b)c(a-b+c)+3ab(a-b+c)
=(a-b+c)(a^2+b^2+c^2-2ab+2ac-2bc-3ac+3...
=(a-b+c)(a^2+b^2+c^2+ab+bc-ca)
=1/2.(a-b+c)(a^2+2ab+b^2+b^2+2bc+c^2+c...
=1/2.(a-b+c)[(a+b)^2+(b+c)^2+(c-a)^2]
\(A=x^2-y^2-x+y\)
\(=\left(x^2-y^2\right)-\left(x-y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)
\(=\left(x+y-1\right)\left(x-y\right)\)
\(B=ax-ab+b-x\)
\(=\left(ax-ab\right)-\left(x-b\right)\)
\(=a\left(x-b\right)-\left(x-b\right)\)
\(=\left(a-1\right)\left(x-b\right)\)
\(D=x^2-2xy+y^2-m^2+2mn-n^2\)
\(=\left(x^2+y^2-2xy\right)-\left(m^2+n^2-2mn\right)\)
\(=\left(x-y\right)^2-\left(m-n\right)^2\)
\(=\left(x-y-m+n\right)\left(x-y+m-n\right)\)
\(E=x^2-y^2-2yz-z^2\)
\(=x^2-\left(y^2+z^2+2yz\right)\)
\(=x^2-\left(y-z\right)^2\)
\(=\left(x+y-z\right)\left(z-y+z\right)\)
\(=>A=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =>A=\left(x-y\right)\left(x+y-1\right)\) ( dấu phía sau bị lỗi nha )
\(=>B=a\left(x-b\right)-\left(x-b\right)\\ =>B=\left(x-b\right)\left(a-1\right)\)
\(=>C=\left(a+b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=>C=3\left(a+b+c\right)\left(x^2+12xy+36y^2\right)\\ =>C=3\left(a+b+c\right)\left(x+6y\right)^2\)
\(\Rightarrow D=\left(x-y\right)^2-\left(m^2-2mn+n^2\right)\\ =>D=\left(x-y\right)^2-\left(m-n\right)^2\)
\(=>D=\left(x-y+m-n\right)\left(x-y-m+n\right)\)
\(=>E=x^2-\left(y^2+2yz+z^2\right)\\ =>E=x^2-\left(y+z\right)^2\)
\(=>E=\left(x-y-z\right)\left(x+y+z\right)\)
T I C K ủng hộ nha
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