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chắc bn nảy hỏi lun cả bài tâp về nhà quá, làm km 1 câu
a) = a+a+a + a +a +1 -a -a -a = a(a+a+1) +(a+a+1) - a(a+a+1)= (a+a+1)(a-a+1)
tự bn thêm mũ 4;3;2 vào được là bn làm dc cac câu sau
\(a.\)
\(\left(x-9\right)^2+12x\left(x-3\right)^2\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)+12x\left(x-3\right)^2\)
\(\Rightarrow\left(x-3\right)\left(x+3+12x+x-3\right)\)
\(\Rightarrow14x\left(x-3\right)\)
\(b.\)
\(a\left(b^2+c^2\right)-b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc\)
\(=ab^2+ac^2-bc^2-ba^2+\left(ca^2+cb^2-2abc\right)\)
\(=ab\left(b-a\right)+c^2\left(a-b\right)+c\left(a-b\right)^2\)
\(=c^2\left(a-b\right)-ab\left(a-b\right)+c\left(a-b\right)^2\)
\(=\left(a-b\right)\left(c^2-ab+ac-bc\right)\)
\(=\left(a-b\right)\left[c\left(c+a\right)-b\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(c-b\right)\left(c+a\right)\)
\(c.\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
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 )3 - x3 - y3 = x3+3x2y+3xy2+y3- x3 - y3 = 3x2y+3xy2 = 3xy( x + y)
b, x3 + y3 + z3 - 3xyz = x3 + 3x2y+3xy2+y3 + z3 - 3x2y-3xy2 -3xyz = (x+y)^3 + z^3 - 3xy( x + y + z)
(x+y+z)[(x+y)^2 - (x+y)z + z^2 ] - 3xy( x + y + z) = (x+y+z) ( x^2 + 2xy + y^2 - xz - yz + z^2 ) - 3xy(x+y+z)
= (x+y+z) ( x^2 + 2xy + y^2 - xz - yz + z^2 - 3xy)
bài tieps theo thì tách từng cái ra rồi rút gọc, còn bnhiu thì đưa 3 ra ngoài
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)
\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)