Question

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)

Ai làm đc mk k nha

Answer 1

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

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

b) Áp dụng đẳng thức ở câu a: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\ge\left(ac+bd\right)^2\)

Dấu "=" xảy ra khi \(\left(ad-bc\right)^2=0\Leftrightarrow ad=bc\)

Answer 2

Link tham khảo : https://hoidap247.com/cau-hoi/165024

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Bài làm : 

a.

(ac + bd)² + (ad – bc)²

= a² c² + 2acbd + b² d² + a² d² - 2adbc + b² c²

= a² c² + b² d² + a² d² + b² c²

= ( a² c² + a² d² ) + ( b² d² + b² c² )

= a² ( c² + d² ) + b² ( c² + d² )

= ( a² + b² ) . ( c² + d² )

Vậy (ac + bd)² + (ad – bc)² = (a² + b²)(c² + d²)

b.

( a² + b² ) . ( c² + d² ) - ( ac + bd )²

= a² c² + a² d² + b² c² + b² d² - a² c²  - 2acbd - b² d²

= a² d² + b² c² - 2acbd

= ( ad )² - 2ad . bc + ( bc )²

= ( ad - bc )² \(\ge\)0

\(\Rightarrow\) (ac + bd)² ≤ (a² + b²)(c² + d²)

Vậy (ac + bd)² ≤ (a² + b²)(c² + d²)

Answer 3

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Answer 4

ta có ĐPCM 

${}^{}+{}^{}=\left({}^{}+{}^{}\right)\left({}^{}+{}^{}\right)$ 

<=> ${}^{}{}^{}+2abcd+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}-2abcd={}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}$

<=> ${}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}={}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}$ (luôn đúng )

Answer 5

b) ta có BĐT cần chứng minh ${}^{}<=\left({}^{}+{}^{}\right)\left({}^{}+{}^{}\right)$

                                               <=> ${}^{}{}^{}+2axby+{}^{}{}^{}<={}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}$

                                              <=> $0<={}^{}{}^{}-2axby+{}^{}{}^{}$

                                              <=> ${}^{}>=0$ (luôn đúng )

Answer 6

ta có ĐPCM 

${}^{}+{}^{}=\left({}^{}+{}^{}\right)\left({}^{}+{}^{}\right)$ 

<=> ${}^{}{}^{}+2abcd+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}-2abcd={}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}$

<=> ${}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}={}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}$

Answer 7

 ta có BĐT cần chứng minh ${}^{}<=\left({}^{}+{}^{}\right)\left({}^{}+{}^{}\right)$

                                               <=> ${}^{}{}^{}+2axby+{}^{}{}^{}<={}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}+{}^{}{}^{}$

                                              <=> $0<={}^{}{}^{}-2axby+{}^{}{}^{}$

                                              <=> ${}^{}>=0$