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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ab+ac}{abc}=2\)
\(\frac{bc+ab+ac}{a+b+c}=2\Leftrightarrow bc+ab+ac=2\left(a+b+c\right)\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}\)( * )
Để \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)thì \(2\left(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}\right)=2\Leftrightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=1\)
\(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{a^2bc+bac^2+ab^2c}{\left(abc\right)^2}=\frac{abc\left(a+b+c\right)}{\left(abc\right)^2}=\frac{a+b+c}{abc}\)
mà a + b + c = abc \(\Rightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{abc}{abc}=1\Leftrightarrow\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\)
thay \(\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\) vào ( * ) ta được \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\left(đpcm\right)\)
\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{bc.ac+ab.ac+ab.bc}{ab.bc.ac}\)
\(=\frac{abc.\left(a+b+c\right)}{a^2b^2c^2}=\frac{a+b+c}{abc}=1\left(\text{vì }a+b+c=abc\right)\)
\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\text{ từ}\left(1\right)\)
Vậy ...
A = \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
= \(a.\frac{a}{b+c}+b.\frac{b}{a+c}+c.\frac{c}{a+b}\)
=\(a.\frac{a}{b+c}+1-1+b.\frac{b}{a+c}+1-1+c.\frac{c}{a+b}+1-1\)
= \(\frac{a\left(a+b+c\right)}{b+c}-a+\frac{b\left(a+b+c\right)}{a+b}-b+\frac{c\left(a+b+c\right)}{a+b}-c\)
= \(\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)
= (a+b+c) - (a+b+c) = 0
Ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=1\)
\(\Leftrightarrow\frac{a+b+c}{abc}=1\Leftrightarrow a+b+c=abc\left(đpcm\right)\)
1/a+1/b+1/c = 0 <=>ab+bc+ca/abc=0
=> ab+bc+ca = 0
Khi đó : (a+b+c)^2 = a^2+b^2+c^2+2.(ab+bc+ca) = a^2+b^2+c^2+2.0 = a^2+b^2+c^2
=> ĐPCM
k mk nha
Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\) (1)
Ta có : a+b+c khác 0
do nếu a+b+c=0=>\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\)=>-3=1(Vô lí)
do a+b+c khác 0 nên ta nhân (a+b+c) vào (1)
=>\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
=>\(\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c\left(a+b\right)+c^2}{a+b}=a+b+c\)
=>\(\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)
=>\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)(ĐPCM)
Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
=> \(\frac{ab+bc+ac}{abc}=0\)
=> \(ab+bc+ac=0\)
Ta lại có \(a+b+c=1\)
=> \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc=1\)
=> \(a^2+b^2+c^2+2\left(ab+ac+bc\right)=1\)
=> \(a^2+b^2+c^2=1\)
Vậy đẳng thức \(a^2+b^2+c^2=0\) là vô lý
cảm ơn b ạ