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\(\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=2a^3-6abc+2b^3+2c^3\)
a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)
\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)
\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)
\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)
\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)
\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)
b) 4a2b2-(a2 +b2-c2)2
=（2ab+a2+b2-c2)(2ab-a2-b2+c2）
=[(a+b)2-c2][c2-(a-b)2]
=(a+b+c)(a+b-c)(c+a-b)(c-a+b)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)
\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)
\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)
\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)
\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)
\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
a: \(9x^2-6x+3\)
\(=\left(9x^2-6x+1\right)+2\)
\(=\left(3x-1\right)^2+2\ge2\)
b: \(6x-x^2+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left(x-3\right)^2+10\le10\)
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
Bài 1 :
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x^2-x\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x^2-x-6\right)\left(x+1\right)\)
\(A=\left(2n-1\right)^3-2n+1\)
\(A=8n^3-6n+6n-1-2n+1\)
\(A=8n^3-2n=2n\left(4n^2-1\right)\)
\(A=2n\left(2n+1\right)\left(2n-1\right)\)
\(A=\left(2n-1\right)2n\left(2n+1\right)⋮6\) ( 3 số tự nhiên liên tiếp)
a) VT = (a+b)(\(a^2-ab+b^2\)) + \(\left(a-b\right)\left(a^2+ab+b^2\right)=a^3+b^3\)\(+a^3-b^3\) = \(2a^3=VP\) (đpcm)
b, VP =\(\left(a+b\right)\left[\left(a-b\right)^2+ab\right]=\left(a+b\right)\left[a^2-2ab+b^2+ab\right]=\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3=VT\left(đpcm\right)\)
c, Ta có : \(VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)(1)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2acbd+b^2d^2+a^2d^2-2adbc+b^2c^2=a^2c^2+b^2d^2+a^2d^2+b^2c^2\) (2)
Từ (1) và (2), ta có \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\left(đpcm\right)\)
a, ta có (a2+ b2)2 = (a2)2 + 2a2b2+ (b2)2( hđt)
= (a2)2 + 2a2b2+(b2)2 - 4a2b2 + 4a2b2
=(a2)2 - 2a2b2 + (b2)2 + 4a2b2
= (a2-b2)2 + 22a2b2 = (a2 -b2) + (2ab)2
vậy .........
b, ta co :(ax +b)2 +(a-bx)2 + c2x2 + c2 = [(ax)2 + 2axb +b2 ] + [ a2 -2abx + (bx)2 ] + (cx)2 + c2 = (ax)2+ (bx)2 + (cx)2 +a2 + b2 +c2 +( 2axb- 2axb)
= x2.(a2+b2+c2) + (a2+b2+c2)
= (x2+1) . (a2+b2+c2)
vay .................
c, Ta có:
\(VP=\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right).c.\left(a+b+c\right)\)
\(=a^3+b^3+3a^2b+3ab^2+c^3+3\left(a+b\right).c.\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3ab.\left(a+b\right)+3\left(a+b\right).c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left[ab+c.\left(a+b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left(ab+ac+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left(ab+ac\right)+\left(cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).a.\left(b+c\right)+c.\left(b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)=VT\)
\(\rightarrow\) đpcm
Chúc bạn học tốt!!!