Question

Chứng minh các hằng đẳng thức sau:

a) \(\left(ax+yy+cz\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)

b) \(\left(ab+bc+ac\right)^2+\left(a^2-bc\right)+\left(b^2-ca\right)^2+\left(c^2-ab\right)^2=\left(a^2+b^2+c^2\right)^2\)

Answer 1

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)