Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\) ⇔ \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=4\)
⇔ \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=1\) ⇔ \(\dfrac{a+b+c}{abc}=1\) ⇔ \(a+b=c=abc\)
Cho biết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\); \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\). CMR: a+b+c=abc
Cho: \(\left(a+b+c\right)^2=a^2+b^2+c^2\) và a, b, c khác 0. CMR: \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Cho: \(\left(a+b+c\right)^2=a^2+b^2+c^2\) và a,b, c khác 0. CMR: \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Cho a+b+c=abc CMR:
\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)=4abc\)
CMR nếu \(\left(a^2-bc\right).\left(b-abc\right)=\left(b^2-ac\right).\left(a-abc\right)\) và các số a, b, c, a-b khác 0 thì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
Cho a,b,c>0, abc\(=\)1
CMR: \(\frac{1}{a^2-ab+b^2}+\frac{1}{b^2-bc+c^2}+\frac{1}{c^2-ca+a^2}\le a+b+c\)
CMR: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{3}{1+abc}\)biết a,b,c\(\ge\)1
10,cho a+b+c=\(\frac{1}{9}\)abc và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\) với a,b,c khác 0.CMR:\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{2}\)