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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, 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 .................
\(\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=2a^3-6abc+2b^3+2c^3\)
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!!!
1. \(\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=a^3+b^3\)
2. \(\left(a-b\right)^3+3ab\left(a-b\right)\)
\(=a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2\)
\(=a^3-b^3\)
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 \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\left(đpcm\right)\)
