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Bài 1"
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\\ =\left(x+2+y\right)\left(x+2-y\right)\)
Baif2:
Có: a+b+c=0
=>a+b=-c
=>\(\left(a+b\right)^3=-c^3\)
=>\(a^3+b^3+3ab\left(a+b\right)=-c^3\)
=>\(a^3+b^3-3abc=-c^3\) (vì a+b=-c)
=>\(a^3+b^3+c^3=3abc\)
Bài 1. Phân tích đa thức thành nhân tử
x2 + 4x - y2 + 4
= ( x2 - y2 ) + ( 4x + 4 )
=( x + y ) ( x - y ) + 4 ( x + 1)
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,\(-4x^2+4x-1\)
\(\Leftrightarrow\left(-2x-1\right)^2\)
b,\(\left(2x+1\right)^2-4\left(x-1\right)^2\)
\(\Rightarrow\left[2x+1-2\left(x-1\right)\right].\left[2x+1+2\left(x-1\right)\right]\)
\(\Rightarrow\left(2x+1-2x+2\right)\left(2x+1+2x-2\right)\)
\(\Rightarrow3\left(4x-1\right)\)
c,\(\left(2x-y\right)^2-4x^2+12x-9\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(4x^2-12x+9\right)\)
\(\Leftrightarrow\left(2x+y\right)^2-\left(2x-3\right)^2\)
\(\Leftrightarrow\left(2x+y-2x+3\right)\left(2x+y+2x-3\right)\)
\(\Rightarrow\left(y+3\right)\left(4x+y-3\right)\)
d,\(\left(x+1\right)^2-4\left(x+1\right)y^2+4y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+2^2y^4\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)2y^2+4\left(y^2\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)-2y^2+\left(2y^2\right)^2\)
\(\Leftrightarrow\left(x+1-2y^2\right)^2\)
Ta có:
\(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+x-2=\left(x-2\right)\left(x^2+x+1\right)\)
Trả lời tội ghê đó bạn nhưng mk gửi một bài mà sao bạn trả lời một câu vậy bạn nhưng dù sao vẫn cảm on nha
a3 + b3 + c3 - 3abc
= (a3 + 3a2b + 3ab2 + b3 ) + c3 - 3abc - 3a2b - 3ab2
=[(a+b)3 + c3 ]- (3abc+3a2b+3ab2)
=(a+b+c)[(a+b)2 - (a+b)c + c2 ] - 3ab(c+a+b)
=(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-bc-ca)
a, \(x^3+x^2-x+2=x^3+2x^2-x^2-2x+x+2\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+x+2\)
\(=\left(x+2\right)\left(x^2-x+1\right)\)
b, \(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)=\left(x+2\right)\left(x^2-8x+15\right)\)
\(a.\)
\(\left(x-9\right)^2+12x\left(x-3\right)^2\)
\(\Rightarrow\left(x-3\right)\left(x+3\right)+12x\left(x-3\right)^2\)
\(\Rightarrow\left(x-3\right)\left(x+3+12x+x-3\right)\)
\(\Rightarrow14x\left(x-3\right)\)
\(b.\)
\(a\left(b^2+c^2\right)-b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc\)
\(=ab^2+ac^2-bc^2-ba^2+\left(ca^2+cb^2-2abc\right)\)
\(=ab\left(b-a\right)+c^2\left(a-b\right)+c\left(a-b\right)^2\)
\(=c^2\left(a-b\right)-ab\left(a-b\right)+c\left(a-b\right)^2\)
\(=\left(a-b\right)\left(c^2-ab+ac-bc\right)\)
\(=\left(a-b\right)\left[c\left(c+a\right)-b\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(c-b\right)\left(c+a\right)\)
\(c.\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a) \(\left(x^2-9\right)^2+12x\left(x-3\right)^2\)
\(=\left[\left(x-3\right)\left(x+3\right)\right]^2+12x\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left(x+3\right)^2+12x\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left[\left(x+3\right)^2+12x\right]\)
\(=\left(x-3\right)^2\left(x^2+6x+3^2+12x\right)\)
\(=\left(x-3\right)^2\left(x^2+18x+9\right)\)
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)