Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/a + 1/b + 1/c = 2
<=> (1/a + 1/b + 1/c) = 4
<=> 1/a^2 1/b^2 + 1/c^2 +2.(1/ab + 1/bc + 1/ca) = 4
<=> 2.(1/ab + 1/bc + 1/ca) = 4-(1/a^2 +1/b^2 + 1/c^2) = 4-2 = 2
<=> 1/ab + 1/bc + 1/ca = 1
<=> a+b+c/abc = 1
<=> a+b+c = abc = a x b x c
Tk mk nha
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) và \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(\Rightarrow\) \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow2^2=\)\(2+2.\left(\frac{a+b+c}{abc}\right)\)
\(\Rightarrow\frac{a+b+c}{abc}=\frac{2^2-2}{2}=0\)
\(\Rightarrow a+b+c=abc\) \(\left(đpcm\right)\)
Ta có
a + b + c = abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
Ta có:a+b+c=abc
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Ta lại có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\Leftrightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=9\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=9\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=9\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=7\)
Làm tương tự bài : Câu hỏi của Hoàng Nguyễn Quỳnh Khanh - Toán lớp 8 - Học toán với OnlineMath