Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

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

\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{d}=1\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=d\end{cases}}\Rightarrow a=b=c=d\)

Ta có: \(VT=a.b^{19}.c^{1999}=d.d^{19}.d^{1999}=d^{2019}=VP\)(đpcm)

thank you bạn gì đó nha

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow ab+bc+ca=0\)

\(a+b+c=\sqrt{2019}\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2019\)

\(\Rightarrow a^2+b^2+c^2=2019\) ( vì \(ab+bc+ca=0\))

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\\ A=a^2+b^2+c^2\\ \Leftrightarrow A=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\\ \Leftrightarrow A=\left(\sqrt{2019}\right)^2-2\cdot0=2019\)

ab² hay là a²b²

Là a.b^2 nhé

Giúp tôi với !!

Lời giải:

Áp dụng TCDTSBN:

$\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}$

$=\frac{a+b+c-d+b+c+d-a+c+d+a-b+d+a+b-c}{d+a+b+c}$

$=\frac{2(a+b+c+d)}{a+b+c+d}=2$
$\Rightarrow a+b+c-d=2d; b+c+d-a=2a; c+d+a-b=2b; d+a+b-c=2c$

$\Rightarrow a+b+c=3d; b+c+d=3a; c+d+a=3b; d+a+b=3c$

Khi đó:

\(P=\frac{a+b+c}{a}.\frac{b+c+d}{b}.\frac{c+d+a}{c}.\frac{a+b+d}{d}\\ =\frac{3d}{a}.\frac{3a}{b}.\frac{3b}{c}.\frac{3c}{d}=81\)

Ta có: \(ac=b^2\)

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

Ta có: \(ab=c^2\)

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

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

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k\)

=>\(\begin{cases}c=ak\\ b=ck=ak\cdot k=ak^2\\ a=bk=ak^2\cdot k=ak^3\end{cases}\Rightarrow\begin{cases}c=ak\\ b=ak^2\\ a\left(k^3-1\right)=0\end{cases}\)

=>\(\begin{cases}c=ak\\ b=ak^2\\ k^3-1=0\end{cases}\Rightarrow\begin{cases}k=1\\ c=a\cdot1=a\\ b=a\cdot1^2=a\end{cases}\)

=>a=b=c

\(P=\frac{a^{555}}{b^{222}\cdot c^{333}}+\frac{b^{555}}{c^{222}\cdot a^{333}}+\frac{c^{555}}{a^{222}\cdot b^{333}}\)

\(=\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}+\frac{a^{555}}{a^{222}\cdot a^{333}}\)

=1+1+1

=3