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a: Sửa đề: x^3-x^2+5x-5
=x^2(x-1)+5(x-1)
=(x-1)(x^2+5)
b: x^3+4x^2+x-6
=x^3-x^2+5x^2-5x+6x-6
=(x-1)(x^2+5x+6)
=(x-1)(x+2)(x+3)
c: \(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
Bài 1 :
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
Bài 2 : Ta có : \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3-3abc=-c^3\) ( Vì \(a+b=-c\) )
\(\Rightarrow a^3+b^3+c^3=3abc\)
Bài 1:
x2 +4x-y2+4
=(x2+4x+4)-y2
=(x+2)2-y2
=(x-y+2)(x+y+2)
Bài 2:
a3+b3+c3 = 3abc
=>a3+b3+c3-3abc=0
=>[(a+b)3+c3]-3ab(a+b)-3abc=0
=>(a+b+c)[(a+b)2-(a+b)c+c2]-3ab(a+b+c)=0
=>(a+b+c)(a2+b2+c2-ac-bc-ab)=0
Từ a+b+c=0
=>0*(a2+b2+c2-ac-bc-ab)=0 (luôn đúng)
\(a.\)
\(\left(x-9\right)^2+12x\left(x-3\right)^2\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)+12x\left(x-3\right)^2\)
\(\Rightarrow\left(x-3\right)\left(x+3+12x+x-3\right)\)
\(\Rightarrow14x\left(x-3\right)\)
\(b.\)
\(a\left(b^2+c^2\right)-b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc\)
\(=ab^2+ac^2-bc^2-ba^2+\left(ca^2+cb^2-2abc\right)\)
\(=ab\left(b-a\right)+c^2\left(a-b\right)+c\left(a-b\right)^2\)
\(=c^2\left(a-b\right)-ab\left(a-b\right)+c\left(a-b\right)^2\)
\(=\left(a-b\right)\left(c^2-ab+ac-bc\right)\)
\(=\left(a-b\right)\left[c\left(c+a\right)-b\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(c-b\right)\left(c+a\right)\)
\(c.\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a) \(\left(x^2-9\right)^2+12x\left(x-3\right)^2\)
\(=\left[\left(x-3\right)\left(x+3\right)\right]^2+12x\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left(x+3\right)^2+12x\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left[\left(x+3\right)^2+12x\right]\)
\(=\left(x-3\right)^2\left(x^2+6x+3^2+12x\right)\)
\(=\left(x-3\right)^2\left(x^2+18x+9\right)\)
a, = (x + y)5 - (x5 + y5)
= (x + y)5 - (x + y)(x4 - x3y + x2y2 - xy3 + y4)
= (x + y) [(x + y)4 - x4 + x3y - x2y2 + xy3 - y4]
= (x + y) (5x3y + 5x2y2 + 5xy3)
= 5xy(x + y)(x2 + xy + y2)
b, = x(x2 - 5xy - 14y2)
= x(x2 - 7xy + 2xy - 14y2)
= x(x + 2y)(x - 7y)
Đặt \(x^2+3x+1=t\)
\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
\(=t\left(t-4\right)-5\)
\(=t^2-4t-5\)
tự làm nốt ý này nhé.
những ý kia lát nx mình làm.
1.\(x^3+6x^2+12xy+8=x^3+3.2x^2+3.2^2x+2^3=\left(x+2\right)^3\)
3.\(x^4+2x^3+x^2-y^2=\left(x^2\right)^2+2x^2.x+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
k mình nha bn !!!!!!! cái 2 bn xem lại đề đi, rồi mình giải cho
1/ \(x^2+2xy+y^2-x-y-12=\left(x+y\right)^2+6\left(x+y\right)+9-7\left(x+y\right)-21\)
\(=\left(x+y+3\right)^2-7\left(x+y+3\right)=\left(x+y+3\right)\left(x+y-4\right)\)
2/ \(4x^4-32x^2+1=\left(4x^4+4x^2+1\right)-36x^2\)
\(=\left(2x^2+1\right)^2-36x^2=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
3/ \(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2=2x^4-2x^3-2x+2\)
\(=2\left(x-1\right)^2\left(x^2+x+1\right)\)
Còn lại tự làm nhé
zễ mà
1,2 áp dụng hằng đẳng thức
3)(2x+y)3
dùng hằng đẳng thức để phân tích:
1) \(\left(a+b\right)^3+\left(a-b\right)^3=\left[\left(a+b\right)+\left(a-b\right)\right]\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2+b^2-a^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
2)\(\left(a+b\right)^3-\left(a-b\right)^3=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2a\left(3a^2+b^2\right)\)
3)\(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)