hơi khó nhưng mong mọi người giải được

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)

\(\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a^{2019}+a^{2019}+a^{2019}}{a^{672}.a^{673}.a^{674}}\)

\(\Rightarrow M=\dfrac{3a^{2019}}{a^{672+673+674}}\)

\(\Rightarrow M=\dfrac{3a^{2019}}{a^{2019}}\)

\(\Rightarrow M=3\)

Có j sai thì mk xl nhé!

Đặt x=a-b; y=b-c; z=c-a

=>x+y+z=0

a+b+c=0

=>a+b=-c; b+c=-a; a+c=-b

Đặt \(X = \frac{a-b}{c} + \frac{b-c}{a} + \frac{c-a}{b}\)

\(=\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc}=\frac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{abc}\)

Đặt \(Y = \frac{c}{a-b} + \frac{a}{b-c} + \frac{b}{c-a}\)

\(=\frac{c}{x}+\frac{a}{y}+\frac{b}{z}=\frac{cyz+axz+bxy}{xyz}\)

Đặt T=cyz+axz+bxy

\(=cyz+(-b-c)xz+bxy=cyz-bxz-cxz+bxy\)

\(=cz(y-x)+bx(y-z)=-3abc\)

Do đó: \(Y = \frac{-3abc}{(a-b)(b-c)(c-a)}\)

A=X*Y

\(=\frac{-\left(a-b\right)\cdot\left(b-c\right)\left(c-a\right)}{abc}\cdot\frac{-3abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

=3

Cứu mình với 9:00 sáng nay mình nộp bài rùi

Lời giải:

$\frac{2022a+b+c}{a}=\frac{a+2022b+c}{b}=\frac{a+b+2022c}{c}$

$=2021+\frac{a+b+c}{a}=2021+\frac{a+b+c}{b}=2021+\frac{a+b+c}{c}$

$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$

$\Rightarrow a+b+c=0$ hoặc $\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$

$\Rightarrow a+b+c=0$ hoặc $a=b=c$

Nếu $a+b+c=0$ thì:

$P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{(-c)}{c}+\frac{(-b)}{b}+\frac{(-a)}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$ thì:

$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$

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

\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)

\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)

a: \(=\dfrac{3}{2}\left(-21-\dfrac{1}{3}+1+\dfrac{1}{3}\right)=\dfrac{3}{2}\cdot\left(-20\right)=-30\)

b: \(=\dfrac{2018}{2019}\left(13-13-\dfrac{2018}{2019}-\dfrac{1}{2019}\right)=-\dfrac{2018}{2019}\)