Cho \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)với mọi x,y ≠ 0 thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
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\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2=0\)\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2abxy-b^2y^2=0\)\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Rightarrow ay-bx=0\)
\(\Leftrightarrow ay=bx\)
\(\Rightarrow\dfrac{a}{x}=\dfrac{y}{b}\)
=> đpcm
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Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Rightarrow\left(ax\right)^2+\left(ay\right)^2+\left(bx\right)^2+\left(by\right)^2=\left(ax\right)^2+2.ax.by+\left(by\right)^2\)
\(\Rightarrow\left(ay\right)^2+\left(bx\right)^2=2.ay.bx\Rightarrow\left(ay\right)^2-2.ay.bx+\left(bx\right)^2=0\)
\(\Rightarrow\left(ay-bx\right)^2=0\Rightarrow ay-bx=0\Rightarrow ay=bx\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\)
Vậy ...
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Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng
Nguyễn Huy Tú Lightning Farron Akai Haruma
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\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
Trừ cả 2 vế cho \(a^2x^2+b^2y^2+c^2z^2\), ta có:
\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)
\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)
\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2z^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)
\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
=> đpcm
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2) ta có: \(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)\) và \(VP=\left(ax+by\right)^2\)
tính hiệu của cả VT và VP
suy ra: \(\left(ay+bx\right)^2=0\Rightarrow ay=bx\)
vì \(x,y\ne0\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\left(đpcm\right)\)
3)(a2+b2+c2)(x2+y2+z2)=(ax+by+cz)2 (1)
biến đổi đẳng thức (1) thành (ay+bx)2 + (bz-cy)2 +(az-cx)2 =0
\(\Rightarrow\) Đpcm
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\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)
Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\) ≥ \(\left(ax+by\right)^2\)
\("="\) ⇔ \(\dfrac{a}{x}=\dfrac{b}{y}\)
Vậy , \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\)\(\left(ax+by\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)