Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{zc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{xbc+yac+zab}{abc}=1\\ \Rightarrow xbc+yac+zab=abc\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=\dfrac{ayz+bxz+cxy}{xyz}=0\\ \Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(abc\right)^2}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc+yac+zab\right)^2}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc\left(ayz+bxz+cxy\right)}\)
\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc.0}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}=1\)
vậy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)(đpcm)
\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.\left(\dfrac{c}{z}+\dfrac{b}{y}+\dfrac{a}{x}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.0=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{abz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{abz+bxz+cxy}{xyz}=0\)
\(\Rightarrow abz+bxz+cxy=0\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\) ( đpcm )
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
Có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
Có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1^2\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(ayz+bxz+cxy=0\right)\)
* Ta có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{axy}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
* Ta có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{b^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)Mà \(cxy+bxz+ayz=0\)
\(\Rightarrow2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
Vậy.........................
Ta có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
=>\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)
=>\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{ayz}{abc}+\dfrac{bxz}{abc}\right)=1\) (1)
Lại có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
=> \(\dfrac{a}{x}.\dfrac{yz}{yz}+\dfrac{b}{y}.\dfrac{xz}{xz}+\dfrac{c}{z}.\dfrac{xy}{xy}=0\)
=>\(\dfrac{ayz}{xuy}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\) (2)
Thay (2) vào (1) ta được
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)
=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
\(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+x^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(x^2+y^2+x^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+x^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2bycz\)\(\Leftrightarrow\left(a^2y^2+2axby+b^2x^2\right)+\left(a^2z^2+2axcz+c^2x^2\right)+\left(b^2z^2+2bycz+c^2y^2\right)=0\)\(\Leftrightarrow\left(ay+bx\right)^2+\left(az+cx\right)^2+\left(bz+cy\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)