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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
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=a^2b^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
b) theo a) \(\Rightarrow\)\(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)
Dấu bằng xảy ra khi ad=bc => a/b=c/d
a,\(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(c^2+d^2\right)\left(a^2+b^2\right)\)
b,Xét hiệu
\(\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ac+bd\right)^2=\left(ad-bc\right)^2\ge0\)
\(\Rightarrow\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có:VT=(ac+bd)2+(ad-bc)2=a2c2+b2d2+2acbd+a2d2+b2c2-2adbc
=a2c2+b2c2+b2d2+a2d2
=(a2+b2)(c2+d2)(ĐPCM)
b)theo câu a) ta có:(ac+bd)2 ≤(a2+b2)(c2+d2)(vì (ad-bc)2 ≥0)
Dấu bằng xảy ra khi:ad=bc
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Bài làm:
a) Ta có: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
=> đpcm
b) CM bất đẳng thức Bunyakovsky chắc được dùng Cauchy đấy nhỉ!
Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
Áp dụng bất đẳng thức Cauchy: \(a^2d^2+b^2c^2\ge2abcd\)
\(\Rightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge a^2c^2+2abcd+b^2d^2=\left(ac+bd\right)^2\)
=> đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2+2abcd-2abcd=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2=0\)
\(\Leftrightarrow0=0\)( luôn đúng )
Vậy \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
b) \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+b^2d^2+2abcd\le a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(\Leftrightarrow a^2d^2+b^2c^2-2abcd\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\)( luôn đúng )
Dấu " = " xảy ra \(\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
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Mấy bài này cứ phá hết ra là xong thôi bạn
\(a,\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2\)
\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)
\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(b,\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2c^2+2abcd+b^2d^2\le a^2c^2+a^2d^2+b^2c^2+b^2d^2\)
\(\Leftrightarrow a^2b^2-2abcd+c^2d^2\ge0\)
\(\Leftrightarrow\left(ab-cd\right)^2\ge0\)(Luôn đúng)
Dấu "=" khi ab = cd
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a) phân tích 2 vế ra là thấy
b)chuyển vế xong phân tích ra chứng minh nó lớn hơn hoặc băng 0 là xong
\(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)
\(=a^2c^2+a^2d^2+b^2d^2+b^2c^2\)
\(=\left(c^2+d^2\right)\cdot\left(a^2+b^2\right)\)
\((ac + bd)^2 + (ad – bc)^2 = (ac)^2 +(bd)^2 + 2(ac)(bd) + (ad)^2 +(bc)^2 - 2(ad)(bc) \)
\( = (ac)^2 +(bd)^2 + (ad)^2 +(bc)^2 + 2abcd – 2abcd\)
\(= a^2c^2 + b^2d^2 + a^2d^2 + b^2c^2\)
\( = (a^2 + b^2)(c^2 + d^2)\)
➤ \((ac + bd)^2 + (ad – bc)^2 = (a^2 + b^2)(c^2 + d^2)\)