Question

[Hạ Vy]

Cho a,b,c>0

Giải pt: \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)

Answer 1

[Buddy]

ĐKXĐ : a;b;c>0;a≠−(b+c);b≠−(c+a);c≠−(a+b)a;b;c≠0;a≠−(b+c);b≠−(c+a);c≠−(a+b)

a+b−xc+b+c−xa+c+a−xb+4xa+b+c=1a+b−xc+b+c−xa+c+a−xb+4xa+b+c=1

⇔(a+b−xc+1)+(b+c−xa+1)+(c+a−xb+1)+4xa+b+c−3−1=0⇔(a+b−xc+1)+(b+c−xa+1)+(c+a−xb+1)+4xa+b+c−3−1=0

⇔a+b+c−xc+a+b+c−xa+a+b+c−xb+4xa+b+c−4=0⇔a+b+c−xc+a+b+c−xa+a+b+c−xb+4xa+b+c−4=0

⇔(a+b+c−x)(1a+1b+1c)+4(x−a−b−c)a+b+c=0⇔(a+b+c−x)(1a+1b+1c)+4(x−a−b−c)a+b+c=0

⇔(a+b+c−x)(1a+1b+1c−4a+b+c)=0⇔(a+b+c−x)(1a+1b+1c−4a+b+c)=0

Do 1a+1b+1c−4a+b+c≠01a+1b+1c−4a+b+c≠0

⇒a+b+c−x=0⇔x=a+b+c⇒a+b+c−x=0⇔x=a+b+c

Vậy ...

Answer 2

[✿✿❑ĐạT̐®ŋɢย❐✿✿]

Ta có pt : \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\) (1)

( ĐK: Do bài cho a,b,c > 0 rồi nên không cần nhé bạn )

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

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

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

Do : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\ne0\forall a,b,c>0\)

Nên : \(a+b+c-x=0\)

\(\Leftrightarrow a+b+c=x\)

Vậy : pt (1) có tập nghiệm \(S=\left\{a+b+c\right\}\)