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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 ...
Ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=1\)
\(\Leftrightarrow\frac{a+b+c}{abc}=1\Leftrightarrow a+b+c=abc\left(đpcm\right)\)
từ a+b+c = 2 suy ra ( a+b+c)^2 =4 <=> a^2 +b^2 +c^2 + 2 (ab+ac+bc)=4 ma2 a^2 + b^2 +c^2 = 2 nên suy ra 2(ab+bc+ac)=2 <=> ab +ac+bc=1 , chia cả 2 vế cho abc khác 0 ta được 1/a+1/b+1/c = 1/abc (đpcm)
Ta có: \(a+b+c=2\)
\(\Leftrightarrow\left(a+b+c\right)^2=4\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=4\)
\(\Leftrightarrow2+2\left(ab+bc+ac\right)=4\)(Vì \(a^2+b^2+c^2=2\))
\(\Leftrightarrow2\left(ab+bc+ac\right)=2\)
\(\Leftrightarrow ab+bc+ac=1\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{abc}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\left(đpcm\right)\)
Ta có : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=4\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)(vì a+b+c=abc)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
(1/a+1/b+1/c)2=1/a^2+1/b^2+1/c^2+2/ab+2/bc+2/ac=1/a^2+1/b^2+1/c^2+2(1/ab+1/ac+1/bc)=1/a^2+1/b^2+1/c^2+2(c/abc+b/abc+a/abc)
=1/a^2+1/b^2+1/c^2+2.1/abc.(a+b+c)
=1/a^2+1/b^2+1/c^2+2.1
=1/a^2+1/b^2+1/c^2+2=4
Suy ra: 1/a^2+1/b^2+1/c^2=4-2=2(điều phải chứng minh)
Ta có : \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\) hay \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=4\)
Mà : \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{\left(a+b+c\right)}{abc}=1\)
Vì : \(\left(a+b+c=abc\right)\)
Nên bằng 1
Vì vậy : \(2\times\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\left(đpcm\right)\)
Bình phương hai vế của 1/a+1/b+1/c=2 ta duoc: 1/a2+1/b2+1/c2+2(a+b+c)/abc=4. Suy ra 1/a2+1/b2+1/c2+2=4(vi a+b+c=abc) Do do 1/a2+1/b2+1/c2=2(dpcm)
chép bài , đồ tồi
Các bạn làm bị sai sai rồi. Chính xác là cái đề sai. Lẽ ra 1/a^2 + 1/b^2 + 1/c^2=3 chứ
Jin Air ban ms la ng sai
Sao nói mấy bạn không chịu tiếp thu vậy? Thấy làm ra kết quả rồi đúng luôn à?
(1/a + 1/b + 1/c)^2= 1/a^2 + 1/b^2 + 1/c^2 + 1/ab + 1/bc + 1/ca
hay 2^2= (1/a^2 + 1/b^2 + 1/c^2) + (1/ab + 1/bc + 1/ca)
=> 4 - (1/ab + 1/bc + 1/ca)=1/a^2 + 1/b^2 + 1/c^2
Mặt khác, 1/ab + 1/bc + 1/ca= c/abc + a/abc + b/abc= c+a+b/abc= 1 (vì c+a+b=abc (gt))
=> 4-1=1/a^2 + 1/b^2 + 1/c^2
3=1/a^2 + 1/b^2 + 1/c^2