Các bạn ơi câu b là bé hơn 2 nhé

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

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

\(\frac{2}{c}=\frac{a}{ab}+\frac{b}{ab}\)

\(\frac{2}{c}=\frac{a+b}{ab}\)

\(2ab=\left(a+b\right).c\)

\(ab+ab=ac+bc\)

\(ab-bc=ac-ab\)

\(b.\left(a-c\right)=a.\left(c-b\right)\)

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

Chọn C.

Phương pháp: Kiểm tra tính đúng sai của từng mệnh đề.

Cách giải:

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{a+b+c}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2.\dfrac{0}{abc}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)

\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)

Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)

\(X=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{c+b}{c}\cdot\frac{c+a}{a}\)

Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\c+b=-a\end{cases}}\)

\(\Rightarrow X=\frac{\left(-a\right)\cdot\left(-b\right)\cdot\left(-c\right)}{abc}=-1\)

nên ta đc X là 1 số nguyên

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow\frac{ab+bc+ac}{abc}=1\Leftrightarrow ab+bc+ac=abc\)

kết hợp gt: a+b+c=1

\(\Rightarrow abc-ab-ac-bc+a+b+c-1=0\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\left(đpcm\right)\)