Đáp án: $Q=\dfrac34$
Giải thích các bước giải:
Ta có:
$\dfrac1a+\dfrac1b+\dfrac1c=2$
$\to \dfrac1a+\dfrac1b=2-\dfrac1c$
$\to (\dfrac1a+\dfrac1b)^2=(2-\dfrac1c)^2$
$\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac2{ab}=4-\dfrac4c+\dfrac1{c^2}$
$\to \dfrac1{a^2}+\dfrac1{b^2}+\dfrac2{ab}-\dfrac1{c^2}=4-\dfrac4c$
$\to \dfrac1{a^2}+\dfrac1{b^2}+4=4-\dfrac4c$
$\to \dfrac1{a^2}+\dfrac1{b^2}=-\dfrac4c$
$\to \dfrac1{a^2}+\dfrac1{b^2}=4\cdot (-\dfrac1c)$
$\to \dfrac1{a^2}+\dfrac1{b^2}=4\cdot (\dfrac1a+\dfrac1b-2)$
$\to \dfrac1{a^2}+\dfrac1{b^2}=4\cdot\dfrac1a+4\cdot\dfrac1b-8$
$\to (\dfrac1{a^2}-\dfrac4a+4)+(\dfrac1{b^2}-\dfrac4b+4)=0$
$\to (\dfrac1a-2)^2+(\dfrac1b-2)^2=0$
$\to \dfrac1a-2=\dfrac1b-2=0$
$\to a=b=\dfrac12$
$\to c=-\dfrac12$
$\to Q=\dfrac34$
