Question

Cho a , b, c khác không ; 1/a - 1/b - 1/c = 1 và a=b+c

chứng minh rằng 1/a^2 + 1/b^2 + 1/c^2=1.

các bạn ơi  giú mình với ''''<_> '''

Answer 1

Từ   \(a=b+c\)   \(\Rightarrow\)  \(a-b-c=0\)    

Ta có:

\(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=1\)

\(\Rightarrow\)  \(\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}\right)^2=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{bc}-\frac{1}{ac}-\frac{1}{ab}\right)=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a}{abc}-\frac{b}{abc}-\frac{c}{abc}\right)=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a-b-c}{abc}\right)=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1\)

\(\Leftrightarrow\)  \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c-c}{abc}\right)=1\)