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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 ...

5 tháng 7 2017

Thanks nhìuyeu

23 tháng 9 2018

Ta có: (a2 + b2)(x2 + y2)

= (ax)2 + a2y2 + b2x2 + (by)2

= (ax + by)2 - 2abxy + a2y2 + b2x2

= (ax + by)2 + (a2y2 + b2x2 - 2abxy)

Mà (a2 + b2)(x2 + y2) = (ax + by)2

\(\Rightarrow\) a2y2 + b2x2 - 2abxy = 0

\(\Rightarrow\) \(\left(ay\right)^2-2.ay.bx+\left(bx\right)^2=0\)

\(\Rightarrow\) \(\left(ay-bx\right)^2=0\)

\(\Rightarrow\) \(ay=bx\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\) (đpcm)

27 tháng 5 2018

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{a}=\dfrac{y}{b}\\\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ay=bx\\bz=cy\\az=cx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ay-bx=0\\bz-cy=0\\az-cx=0\end{matrix}\right.\)

\(\Leftrightarrow\left(ax-by\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

\(\Leftrightarrow\left(a^2x^2-2axby+b^2y^2\right)+\left(b^2z^2-2bzcy+c^2y^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)=0\)

\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2-\left(a^2x^2+b^2b^2+c^2y^2+2axby+2azcx+2bzcy\right)=0\)

\(\Leftrightarrow x^2\left(a^2+b^2+c^2\right)+y^2\left(a^2+b^2+c^2\right)+z^2\left(a^2+b^2+c^2\right)-\left(ax+ab+cz\right)^2=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)-\left(ax+by+cz\right)^2=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

27 tháng 5 2018

Ta có : \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\) ( theo bđt Bu-nhi-a Cop-xki )

Dấu "=" xảy ra khi \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

Vậy nếu \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) thì \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

28 tháng 7 2017

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2=0\)\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2abxy-b^2y^2=0\)\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\)

\(\Rightarrow ay-bx=0\)

\(\Leftrightarrow ay=bx\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{y}{b}\)

=> đpcm

28 tháng 7 2017

\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)

\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

Trừ cả 2 vế cho \(a^2x^2+b^2y^2+c^2z^2\), ta có:

\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)

\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)

\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2z^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)

\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)

\(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

=> đpcm

3 tháng 10 2016

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\Leftrightarrow ay=bx\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)

3 tháng 10 2016

Cảm ơn bạn nhiều

4 tháng 1 2018

\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)

Ta co

\(\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2-2bcyz+bcz^2+acz^2-2aczx+acx^2+abx^2-2abxy+aby^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2\left(axby+bcyz+axcz\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{\left(acx^2+abx^2+a^2x^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(c^2z^2+acz^2+bcz^2\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\dfrac{1}{a+b+c}\) ( dpcm)