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

\(\dfrac{a}{3b}=\dfrac{b}{3c}=\dfrac{c}{3d}=\dfrac{d}{3a}=\dfrac{a+b+c+d}{3b+3c+3d+3x}=\dfrac{a+b+c+d}{3.\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow a=\dfrac{1}{3}.3b=b\\ \Rightarrow b=\dfrac{1}{3}.3c=c\\ \Rightarrow c=\dfrac{1}{3}.3d=d\\ \Rightarrow d=\dfrac{1}{3}.3a=a\) 

➩\(\text{a=b=c=d}\)

Tick cho mình nhé

\(a^3+b^3+c^3=3abc\\ \left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\\ \left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\\ \left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Do \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ \left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Rightarrow a=b=c\)

=>P=2009³

Lời giải:

PT $\Leftrightarrow \frac{a+b-x}{c}+1+\frac{a+c-x}{b}+1+\frac{b+c-x}{a}+1+\frac{4x}{a+b+c}-4=0$

$\Leftrightarrow \frac{a+b+c-x}{c}+\frac{a+b+c-x}{b}+\frac{a+b+c-x}{a}-\frac{4(a+b+c-x)}{a+b+c}=0$

$\Leftrightarrow (a+b+c-x)(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c})=0$

$\Rightarrow a+b+c-x=0$ hoặc $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$
Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}=0$, khi đó $x$ nhận mọi giá trị thực.

Nếu $\frac{1}{c}+\frac{1}{b}+\frac{1}{a}-\frac{4}{a+b+c}\neq 0$

$\Rightarrow a+b+c-x=0$
$\Rightarrow x=a+b+c$

Lời giải:
Bạn tham khảo cách làm tương tự tại đây:

https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584

Kết quả $P=8$ hoặc $P=-1$

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Có:

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

\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right).\left(a+b+c\right)=1.\left(a+b+c\right)\)

\(\Rightarrow\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{a+c}+\dfrac{c\left(a+b+c\right)}{a+b}=a+b+c\)

\(\Rightarrow\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{a+c}+\dfrac{c^2+c\left( a+b\right)}{a+b}=a+b+c\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c\left(a+b\right)}{a+b}=a+b+c\)

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

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

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)

j giàu thế :) cho xin ít đi

\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{2a+2b+2c}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c=a+5\\a+b+c=b-2\\a+b+c=c-3\end{matrix}\right.\)

Lại có \(\dfrac{1}{a+b+c}=2\Rightarrow a+b+c=\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}a+5=\dfrac{1}{2}\\b-2=\dfrac{1}{2}\\c-3=\dfrac{1}{2}\end{matrix}\right.\)

Từ đó tự giải ra

Áp dụng t/c dtsbn:

\(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{b+c-5+a+c+2+a+b+3}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}b+c-5=2a\\a+c+2=2b\\a+b+3=2c\end{matrix}\right.\)\(\left(1\right)\)

Mặt khác \(\dfrac{1}{a+b+c}=\dfrac{b+c-5}{a}=2\)\(\Rightarrow a+b+c=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{2}-c\\a+c=\dfrac{1}{2}-b\\b+c=\dfrac{1}{2}-a\end{matrix}\right.\)\(\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-a-5=2a\\\dfrac{1}{2}-b+2=2b\\\dfrac{1}{2}-c+3=2c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{3}{2}\\b=\dfrac{5}{6}\\c=\dfrac{7}{6}\end{matrix}\right.\)

\(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)=\left(-\dfrac{3}{2}-3.\dfrac{5}{6}\right)\left(\dfrac{5}{6}-\dfrac{7}{6}\right)\left(3.\dfrac{7}{6}+\dfrac{3}{2}\right)=\dfrac{20}{3}\)