quy đồng lên ta có bc/abc+ac/abc+ab/abc=0
bc+ac+ab/abc=0
suy ra bc+ac+ab=0
quy đồng M ta có (b+c)bc/abc+(c+a)ac/abc+(a+b)ab/abc
=(b^2c+bc^2+ac^2+a^2c+a^2b+ab^2)/abc
=(b^2c+ab^2+abc+bc^2+ac^2+abc+a^2c+a^2b+abc-3abc)/abc
=(b(bc+ab+ac)+c(bc+ac+ab)+a(ac+ab+bc)-3abc)/abc
=-3abc/abc=-3
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Cho a,b,c >0 thỏa mãn a+b+c\(\le\)\(\frac{3}{2}\).Chứng minh
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)6
b,a+ b+ c+ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)\(\ge\)\(\frac{15}{2}\)
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=C\) a nhân b nhân c khác 0
Tinh E = \(\frac{ab}{c^2}+\frac{ac}{b^2}+\frac{bc}{a^2}\)
Cho a,b,c khác 0 và a+b+c khác 0 thỏa mãn\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\). Chứng minh rằng \(\frac{1}{a^{2011}}+\frac{1}{b^{2011}}+\frac{1}{c^{2011}}=\frac{1}{a^{2011}+b^{2011}+c^{2011}}\)
Cho a+b+c=0. Tinh Q=\((\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b})(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a})\)
cho abc khác 0 tm:a+b+c khác 0 và\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
CMR:\(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{a^{2005}+b^{2005}+c^{2005}}\)
cho cc số a;b; thỏa mãn a+b+c khac 0 va\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\) khi đó giá trị của M=\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\)?
