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tách ít ít ra thôi. để cả cộp thế này k ai làm cho đâu. mệt quá
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
Trừ cả hai vế cho \(a^2x^2+b^2y^2+c^2z^2\), có :
\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)
\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)
\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2x^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)
\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Mà \(\hept{\begin{cases}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}ay=bx\\az=cx\\bz-cy\end{cases}}\)
\(\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Vậy ...
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2=a^2k^2\\y^2=b^2k^2\\z^2=c^2k^2\end{matrix}\right.\)
Ta có: \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)\)
\(=\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)\)
\(=\left(a^2+b^2+c^2\right)^2\cdot k^2\)(1)
Ta có: \(\left(ax+by+cz\right)^2\)
\(=\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2\)
\(=\left(a^2k+b^2k+c^2k\right)^2\)
\(=\left(a^2+b^2+c^2\right)^2\cdot k^2\)(2)
Từ (1) và (2) suy ra \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)(đpcm)
Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow\left\{{}\begin{matrix}x=ak\\y=bk\\z=ck\end{matrix}\right.\)
Ta có:
\(\left\{{}\begin{matrix}\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)\\\left(a.ak+b.bk+c.ck\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k^2\left(a^2+b^2+c^2\right)^2\\\left(a^2k+b^2k+c^2k\right)^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k^2\left(a^2+b^2+c^2\right)^2\\\left[k\left(a^2+b^2+c^2\right)\right]^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}k^2\left(a^2+b^2+c^2\right)^2\\k^2\left(a^2+b^2+c^2\right)^2\end{matrix}\right.\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)
Vậy......................(đpcm)
Chúc bạn học tốt!!!