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ta có: x/a = y/b =z/c =xa/a^2 =yb/b^2 =zc/c^2 = (ax+by+cz)/(a^2+b^2+c^2)
=>x/a = (ax+by+cz)/(a^2+b^2+c^2) (1)
mặt khác ta có: x/a=y/b=z/c <=> x^2/a^2 =y^2/b^2 =z^2/c^2 = (x^2+y^2+z^2 ) / (a^2+b^2+c^2)
=>x^2/a^2 = (x^2+y^2+z^2 ) / (a^2+b^2+c^2) (2)
từ (1) và (2) ta => (ax+by+cz)^2/(a^2+b^2+c^2)^2 = (x^2+y^2+z^2 ) / (a^2+b^2+c^2)
=> (x^2+y^2+z^2).(a^2+b^2+c^2)=(ax+by+cz)^2 => đpcm
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)
\(\dfrac{bz-cy}{a}=\dfrac{b.ck-c.bk}{a}=\dfrac{0}{a}=0\)(1)
\(\dfrac{cx-az}{b}=\dfrac{c.ak-a.ck}{b}=\dfrac{0}{b}=0\)(2)
\(\dfrac{ay-bz}{c}=\dfrac{a.bk-b.ak}{c}=\dfrac{0}{c}=0\)(3)
từ (1),(2) và(3) suy ra \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\left(đpcm\right)\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Theo tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8x+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{x}{6}=\dfrac{z}{12}\\\dfrac{y}{6}=\dfrac{z}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
Kết luận ...
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}=3\)
\(\Rightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}=0\)
\(\Rightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)
Thay vào \(2x^2+3y^2=30\) ta được:
\(2.\left(3k\right)^2+3.\left(4k\right)^2=30\)
\(\Leftrightarrow66k^2=30\)
\(\Rightarrow k^2=\dfrac{5}{11}\)
\(\Rightarrow k=\pm\dfrac{\sqrt{55}}{11}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt{55}}{11};y=\dfrac{4\sqrt{55}}{11}\\x=-\dfrac{3\sqrt{55}}{11};y=-\dfrac{4\sqrt{55}}{11}\end{matrix}\right.\)