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a: \(\left(a^2-b^2\right)^2+\left(2ab\right)^2\)
\(=a^4-2a^2b^2+b^4+4a^2b^2\)
\(=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2\)
b: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
c: \(\left(ax+b\right)^2+\left(a-bx\right)^2+c^2x^2\)
\(=a^2x^2+b^2+a^2+b^2x^2+c^2x^2\)
\(=a^2\left(x^2+1\right)+b^2\left(x^2+1\right)+c^2x^2\)
\(=\left(x^2+1\right)\left(a^2+b^2\right)+c^2x^2\)
a) Sửa đề: \(\left(ax+by+cx\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2\)
= a2x2 + b2y2 + c2x2 + 2axby + 2bycz + 2axcz + b2x2 - 2bxay + a2y2 + c2y2 - 2cybz + b2z2 + a2z2 - 2azcx + c2x2
= a2x2 + b2y2 + c2x2 + b2x2 + a2y2 + c2y2 + b2z2 + a2z2 + c2x2
= a2(x2+y2+z2) + b2(x2+y2+z2) + c2(x2+y2+z2)
= (a2+b2+c2)(x2+y2+z2) (đpcm)
b) Đặt x = b; y = c; z = a, ta có:
\(\left(ay+bz+cx\right)^2+\left(az-by\right)^2+\left(bx-cz\right)^2+\left(cy-ax\right)^2\)
= a2y2 + b2z2 + c2x2 + 2aybz + 2bzcx + 2aycx + a2z2 - 2azby + b2y2 + b2x2 - 2bxcz + c2z2 + c2y2 - 2cyax + a2x2
= a2y2 + b2z2 + c2x2 + a2z2 + b2y2 + b2x2 + c2z2 + c2y2 + a2x2
= (a2+b2+c2)(x2+y2+z2)
Thay b = x, c = y, a = z, ta có:
(a2+b2+c2)(x2+y2+z2) = (a2+b2+c2)2 (đpcm)
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))
Nó là bđt bunyakovsky luôn rồi mà bạn,lên google sẽ có cách chứng minh