Question

[Mạnh]

a) Chứng minh: (ac + bd)2 + (ad – bc)2 = (a2 + b2)(c2 + d2) b) Chứng minh bất dẳng thức Bunhiacôpxki: (ac + bd)2 ≤ (a2 + b2)(c2 + d2) giúp nha

Answer 1

[Lại Thanh Tùng]

hok

tốt 

nha

Answer 2

[Kimmdayy=3]

a) Ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$$=a^2{c}^2+2acbd+b^2{d}^2+a^2{d}^2-2adbc+b^2{c}^2$

$=\left(a^2{c}^2+b^2{c}^2\right)+\left(a^2{d}^2+b^2{d}^2\right)$$=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)$$=\left(a^2+b^2\right)\left(c^2+d^2\right)$

b) Ta có $0\le {\left(ad-bc\right)}^2$$\iff {\left(ac+bd\right)}^2\le {\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$

Mà theo câu a, ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2=\left(a^2+b^2\right)\left(c^2+d^2\right)$

Nên ${\left(ac+bd\right)}^2\le \left(a^2+b^2\right)\left(c^2+d^2\right)$