\(\Leftrightarrow\dfrac{x-a-b-c}{b+c}+\dfrac{x-b-a-c}{a+c}+\dfrac{x-c-a-b}{a+b}=0\)

\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=a+b+c\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=0\end{matrix}\right.\)

Xét \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=0\)

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

\(\Rightarrow\left(a+b\right)\left(b+c\right)+\left(c+a\right)\left(b+c\right)+\left(a+b\right)\left(a+c\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2+3\left(ab+bc+ca\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=0\\ab+bc+ca=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\ab-\left(a+b\right)b-\left(a+b\right)a=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\ab+a^2+b^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c=0\)

Vậy với x=a+b+c hoặc a=b=c=0 thì pt thỏa mãn.

Nâng cao và pt tập 2 

\(PT\Leftrightarrow\dfrac{x-a}{b+c}-1+\dfrac{x-b}{c+a}-1+\dfrac{x-c}{a+b}-1=\dfrac{3x}{a+b+c}-3\)

\(\Leftrightarrow\dfrac{x-a-b-c}{b+c}+\dfrac{c-a-b-c}{c+a}+\dfrac{x-a-b-c}{a+b}=\dfrac{3\left(x-a-b-c\right)}{a+b+c}\)

\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}-\dfrac{3}{a+b+c}\right)=0\)

Nếu \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}-\dfrac{3}{a+b+c}=0\) thì PT có nghiệm với mọi \(x\in R\)

Nếu \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}-\dfrac{3}{a+b+c}\ne0\) thì PT có nghiệm là \(x=a+b+c\)

bạn có thể cho mình cách gải được k

cách giải là đi hỏi thầy,cô

Lời giải:
PT $\Leftrightarrow 3x-\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}\right)=a+b+c$

$\Leftrightarrow 3x=\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}+a+b+c$

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

$\Leftrightarrow x=\frac{1}{3}(ab+bc+ac)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$

\(\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)=0\\ \Leftrightarrow x^2-ax-bx+ab+x^2-bx-cx+bc+x^2-cx-ax+ac=0\\ \Leftrightarrow3x^2-2\left(a+b+c\right)x+ab+bc+ca=0\left(1\right)\)

pt(1) là pt bậc 2 ẩn x có:

\(\Delta'=\left(-a-b-c\right)^2-3\left(ab+bc+ca\right)\\ =a^2+b^2+c^2+2ab+2bc+2ca-3\left(ab+bc+ca\right)\\ =a^2+b^2+c^2-ab-bc-ca\\ =\dfrac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

pt có no kép nên delta' =0

nên: \(\dfrac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\\ \Rightarrow a-b=b-c=c-a=0\\ \Rightarrow a=b=c\)

bonus: khi đó pt: \(3\left(x-a\right)^2=0\Leftrightarrow x-a=0\Leftrightarrow x=a\)

=> x=a=b=c

ĐKXĐ: \(x\ge1\)

\(x-1+\sqrt{5+\sqrt{x-1}}=5\)

Đặt \(\sqrt{x-1}=t\ge0\)

\(\Rightarrow t^2+\sqrt{t+5}=5\)

Đặt \(\sqrt{t+5}=u>0\Rightarrow u^2-t=5\)

\(\Rightarrow t^2+u=u^2-t\Leftrightarrow t^2-u^2+t+u=0\)

\(\Leftrightarrow\left(t+u\right)\left(t-u+1\right)=0\)

\(\Leftrightarrow t-u+1=0\) (do \(t>0;u>0\Rightarrow t+u>0\))

\(\Leftrightarrow t+1=\sqrt{t+5}\)

\(\Leftrightarrow t^2+2t+1=t+5\Leftrightarrow t^2+t-4=0\)

\(\Rightarrow t=\dfrac{-1+\sqrt{17}}{2}\)

\(\Rightarrow x=t^2+1=\dfrac{11-\sqrt{17}}{2}\)