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
b) Làm bừa :|
Xét hiệu: \(\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ac+bd\right)^2\ge0\)
\(\Leftrightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2-\left(ac+bd\right)^2\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\left(TM\right)\)
Vậy: ..... :|
b)Ta có:\(\left(ac+bd\right)^2\le\left(a^2+c^2\right)\left(b^2+d^2\right)\)
\(\Leftrightarrow a^2b^2+c^2d^2+2abcd\le a^2b^2+a^2d^2+b^2c^2+c^2d^2\)
\(\Leftrightarrow a^2d^2+b^2c^2-2abcd\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\forall a,b,c,d\in R\)
Đẳng thức xảy ra khi \(\left(ad-bc\right)^2=0\)\(\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
a, (ac + bd)2 + (ad - bc)2=(a2 + b2)(c2 + d2)
↔ (ac)2 + 2abcd + (bd)2 + (ad)2- 2abcd + (bc)2 = (ac)2 + (bc)2 + (ad)2 + (bd)2
↔ (ac)2 + (bd)2 + (ad)2 + (bc)2 = (ac)2 + (bc)2 + (ad)2 + (bd)2 ( luôn đúng )
b, Mình k viết được dấu lớn hơn hoặc bằng nên mình không viết . Bạn cứ biến đổi tương đương là ra
α
a) Ta có
\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow\left(ac\right)^2+2acbd+\left(bd\right)^2+\left(ad\right)^2-2adbc+\left(bc\right)^2=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(\Leftrightarrow\left(ac\right)^2+\left(bd\right)^2+\left(ad\right)^2+\left(bc\right)^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\) (luôn luôn đúng)
\(\Rightarrow\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
b) Ta có
\(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow\left(ac\right)^2+2acbd+\left(bd\right)^2\le c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2c^2+2acbd+b^2d^2\le a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Leftrightarrow2acbd\le b^2c^2+a^2d^2\)
\(\Leftrightarrow-\left(b^2c^2\right)+2acbd-a^2d^2\le0\)
\(\Leftrightarrow-\left(b^2c^2-2acbd+a^2d^2\right)\le0\)
\(\Leftrightarrow-\left(bc-ad\right)^2\le0\) ( điều phải chứng minh )
