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\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\\ \Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\\ \Rightarrow ayz+bxz+cxy=0\left(1\right)\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\\ \Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{2\left(cxy+bxz+ayz\right)}{abc}=1\left(2\right)\)
Từ 1 và 2 suy ra điều cần chứng minh
Ta có:
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\left(1\right)\)
Mặt khác:
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ac}\right)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{cxy+ayz+bzx}{abc}=1\left(2\right)\)
Từ (1) và (2) ta có đpcm.
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), suy ra \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\frac{xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\) \(\left(1\right)\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) (theo gt) nên từ \(\left(1\right)\) \(\Rightarrow\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\) \(\left(đpcm\right)\)
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He he...
Ta có:
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)
Ta có:
\(x+y+z=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\).
\(x+y+z=\frac{\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(a+b\right)\left(c+a\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Ta có:
\(\left(a-b\right)\left(b+c\right)\left(c+a\right)+\left(b-c\right)\left(c+a\right)\left(a+b\right)+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).
\(=\left(c+a\right)\left[\left(a-b\right)\left(b+c\right)+\left(b-c\right)\left(a+b\right)\right]+\left(c-a\right)\left(a+b\right)\left(b+c\right)\).
\(=\left(c+a\right)\left(ab+ac-b^2-bc+ab+b^2-ac-bc\right)\)\(+\left(c-a\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(c+a\right)\left(2ab-2bc\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=2b\left(c+a\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(2bc+2ab\right)\left(a-c\right)-\left(a-c\right)\left(ab+ac+b^2+bc\right)\).
\(=\left(a-c\right)\left(2ab+2bc-ab-ac-b^2-bc\right)\).
\(=\left(a-c\right)\left(ab+bc-b^2-ac\right)=\left(a-c\right)\left[\left(ab-b^2\right)-\left(ac-bc\right)\right]\).
\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]=\left(a-c\right)\left(a-b\right)\left(b-c\right)\).
Do đó\(x+y+z=\frac{\left(a-c\right)\left(a-b\right)\left(b-c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\).
Mà \(xyz=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)nên:
\(x+y+z=-xyz\).
\(\Rightarrow x+y+z+xyz=0\)(điều phải chứng minh).