toán 9

V

vocamtu

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bigbang195

Cho [TEX]a+b+c =0[/TEX] và [TEX]a,b,c[/TEX]khác [TEX]0[/TEX]. Chứng minh rằng đẳng thức:


[TEX]\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}[/TEX]=[TEX]\|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\|\[/TEX]


Bài này chứng minh đơn thôi cậu à

[TEX]\Leftrightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2[/TEX]

đúng vì

[TEX]\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{a+b+c}{abc}=0[/TEX]
 
P

panh29

Ta có : [TEX](\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac})[/TEX]
Mà : a+b+c=0\Rightarrow [TEX]\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{c+a+b}{abc}=0[/TEX]\Rightarrow [TEX](\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\Rightarrow |\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}[/TEX]
 
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