Question

a)Cho 3 số a,b,c thỏa mãn abc=2019. Tính giá trị biểu thức:

M=\(\frac{2019a}{ab+2019a+2019}+\frac{b}{bc+b+2019}+\frac{c}{ac+c+1}\)

b)Cho b,c ≠0 và a+b+c=abc và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

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

Answer 1

Có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{cb}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{a+b+c}{abc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{abc}{abc}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

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

đpcm

\(M=\frac{2019a}{ab+2019a+2019}+\frac{b}{bc+b+2019}+\frac{c}{ca+c+1}\)

\(M=\frac{abc.a}{ab+abc.a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ca+c+1}\)

\(M=\frac{ca}{1+ca+c}+\frac{1}{c+1+ac}+\frac{c}{ca+c+1}\)

\(M=\frac{ca+a+1}{1+ca+c}\)

\(M=1\)