Question

[Kaya Renger]

Với a,b,c dương , giải phương trình 

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

Answer 1

[Riio Riyuko]

pt <=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x+a+b+c}{a+b+c}=5\) (Cộng 4 vào mỗi vế)

   <=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x+a+b+c-5\left(a+b+c\right)}{a+b+c}=0\)

   <=>  \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4a-4b-4c}{a+b+c}=0\)

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

Áp dụng bất đẳng thức Cauchy - Schwarz dạng engel, ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}>\frac{4}{a+b+c}\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}>0\)

Vậy phương trình trên có nghiệm là 

x = a + b + c 

Answer 2

[nguyen duong lam]

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