Cho abc(a+b+c) khác 0. Giải phương trình ẩn x:

(x-a)/bc+(x-b)/ac+(x-c)/ab=1/2(1/a+1/b+1/c)

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bf

gợi ý nha (mik lm còn j là hok nx )   (x+a)(x+b)(x+c)=x2+(a+b+c)x2+(ab+bc+ac)x+abc

            Muốn chứng minh được ta phải chứng minh vế trái    

(x2+bx+ax+ab)(x+c)=x3+ax2+bx2+cx2+abx+bcx+acx+abc

     x3+ax2+bx2+cx2+abx+bcx+acx+abc=x3+ax2+bx2+cx2+abx+bcx+acx+abc(1)

Vì hai biểu thức trên (1) giông nhau

               Do đó (x+a)(x+b)(x+c)=x2+(a+b+c)x2+(ab+bc+ac)x+abc

Lời giải:

\(\frac{x-b-c}{a}+\frac{x-a-c}{b}+\frac{x-a-b}{c}=3\)

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

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

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

Vì $abc(ab+bc+ac)\neq 0\Rightarrow \frac{ab+bc+ac}{abc}\neq 0$

$\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\neq 0(2)$

Từ $(1);(2)\Rightarrow x-a-b-c=0\Rightarrow x=a+b+c$

\(a)\) ĐKXĐ: \(a\ne-b;a\ne-c;b\ne-c\)

\(\dfrac{x-ab}{a+b}+\dfrac{x-ac}{a+c}+\dfrac{x-bc}{b+c}=a+b+c\)

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

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

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

Vì \(a,b,c>0\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}>0\)

\(\Leftrightarrow x-ab-ac-bc=0\)

\(\Leftrightarrow x=ab+ac+bc\)

Đáp án: D