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Điều kiện a; b ; c khác 0
\(\Rightarrow x.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=2.\left(\frac{bc+ac+ab}{abc}\right)+\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\)
\(\Rightarrow x.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=\frac{2bc+2ac+2ab}{abc}+\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}\)
\(\Rightarrow x.\left(\frac{a+b+c}{abc}\right)=\frac{\left(a+b+c\right)^2}{abc}\)
\(\Rightarrow x.\left(a+b+c\right)=\left(a+b+c\right)^2\)
Nếu a+ b+ c khác 0 => phương trình có nghiệm duy nhất là \(\Rightarrow x=a+b+c\)
Nếu a+ b + c = 0 => x. 0 = 0 => pt có vô số nghiêm
a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4\left(a+b+c\right)}{a+b+c}=0\)
\(\Leftrightarrow\left(x-a-b-x\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
b)đề bài như trên
\(\Leftrightarrow\left(\frac{x-a-b-c}{bc}\right)+\left(\frac{x-b}{ca}-\frac{1}{a}-\frac{1}{c}\right)+\left(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b}\right)=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{x-a}{bc}+\frac{x-b}{ac}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\frac{ax-a^2+bx-b^2+cx-c^2}{abc}=2\left(\frac{ab+bc+ac}{abc}\right)\)
\(ax-a^2+bx-b^2+cx-c^2=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)-\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(x\left(a+b+c\right)=a^2+b^2+c^2+2ab+2bc+2ac\)
\(x=a+b+c\)
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 $(1);(2)\Rightarrow x-a-b-c=0\Rightarrow x=a+b+c$
Câu 2/
\(\frac{a^2+bc}{a^2\left(b+c\right)}+\frac{b^2+ca}{b^2\left(c+a\right)}+\frac{c^2+ab}{c^2\left(a+b\right)}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{a^2+bc}{a^2\left(b+c\right)}-\frac{1}{a}+\frac{b^2+ca}{b^2\left(c+a\right)}-\frac{1}{b}+\frac{c^2+ab}{c^2\left(a+b\right)}-\frac{1}{c}\ge0\)
\(\Leftrightarrow\frac{\left(b-a\right)\left(c-a\right)}{a^2\left(b+c\right)}+\frac{\left(a-b\right)\left(c-b\right)}{b^2\left(c+a\right)}+\frac{\left(a-c\right)\left(b-c\right)}{c^2\left(a+b\right)}\ge0\)
\(\Leftrightarrow a^4b^4+b^4c^4+c^4a^4-a^4b^2c^2-a^2b^4c^2-a^2b^2c^4\ge0\)
\(\Leftrightarrow a^4b^4+b^4c^4+c^4a^4\ge a^4b^2c^2+a^2b^4c^2+a^2b^2c^4\left(1\right)\)
Ma ta có: \(\hept{\begin{cases}a^4b^4+b^4c^4\ge2a^2b^4c^2\left(2\right)\\b^4c^4+c^4a^4\ge2a^2b^2c^4\left(3\right)\\c^4a^4+a^4b^4\ge2a^4b^2c^2\left(4\right)\end{cases}}\)
Cộng (2), (3), (4) vế theo vế rồi rút gọn cho 2 ta được điều phải chứng minh là đúng.
PS: Nếu nghĩ được cách khác đơn giản hơn sẽ chép lên cho b sau. Tạm cách này đã.