a) Chứng minh: (ac + bd) 2 + (ad – bc) 2 = (a 2 + b 2 )(c 2 + d 2 ) b) Chứng minh bất dẳng thức Bunhiacôpxki: (ac + bd) 2 ≤...

a) Cách lầy lội nhất khai triển hết ra :| \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\) \(=\left(a^2c^2+b^2c^2\right)+\left(b^2d^2+a^2d^2\right)=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\) Câu trả lời: a) Cách lầy lội nhất khai triển hết ra :| \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\) \(=\left(a^2c^2+b^2c^2\right)+\left(b^2d^2+a^2d^2\right)=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\) \(\Leftrightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2\) Biến đổi vế traias ta có: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\) \(=a^2c^2+b^2d^2+a^2d^2+b^2c^2=VP\) =>đpcm b)Có: \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\) \(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le a^2c^2+a^2d^2+b^2c^2+b^2d^2\) \(\Leftrightarrow-a^2d^2+2abcd-b^2c^2\le0\) \(\Leftrightarrow-\left(a^2d^2-2abcd+b^2c^2\right)\le0\) \(\Leftrightarrow-\left(ad-bc\right)^2\le0\) , luôn luôn đúng =>đpcm Câu trả lời: a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\) \(\Leftrightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2\) Biến đổi vế traias ta có: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\) \(=a^2c^2+b^2d^2+a^2d^2+b^2c^2=VP\) =>đpcm b)Có: \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\) \(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le a^2c^2+a^2d^2+b^2c^2+b^2d^2\) \(\Leftrightarrow-a^2d^2+2abcd-b^2c^2\le0\) \(\Leftrightarrow-\left(a^2d^2-2abcd+b^2c^2\right)\le0\) \(\Leftrightarrow-\left(ad-bc\right)^2\le0\) , luôn luôn đúng =>đpcm