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

x=a+b+c

x vô số nghiệm nếu a+b+c khác 0

x=a+b+c nếu a+b+c = 0

\(ĐKXĐ:a,b,c\ne0\)

\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)

\(\Leftrightarrow\frac{xa-a^2}{abc}+\frac{xb-b^2}{abc}+\frac{xc-c^2}{abc}=\frac{2bc}{abc}+\frac{2ac}{abc}+\frac{2ab}{abc}\)

\(\Leftrightarrow\frac{xa-a^2+xb-b^2+xc-c^2}{abc}=\frac{2bc+2ac+2ab}{abc}\)

\(\Leftrightarrow xa-a^2+xb-b^2+xc-c^2=2bc+2ac+2ab\)

\(\Leftrightarrow xa+xb+xc=2bc+2ac+2ab+a^2+b^2+c^2\)

\(\Leftrightarrow x\left(a+b+c\right)=\left(a+b+c\right)^2\)

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

Vậy x = a + b + c

\(ĐKXĐ:a,b,c\ne0\)

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

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

\(\Leftrightarrow1+\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}=4\)

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

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

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

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

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

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

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

+) Nếu \(\Rightarrow\left(a+b+c-x\right)=0\)thì x = a + b + c

+) Nếu \(\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)thì x thỏa mãn với mọi số

iu cậu quáaa <3

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

1.

\(P=\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\)

\(P^2=\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}+2a^2+2b^2+2c^2\)

Áp dụng BĐT Cô-si :

\(P^2=\frac{1}{2}\cdot\left(\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+\frac{b^2c^2}{a^2}+\frac{a^2b^2}{c^2}+\frac{a^2b^2}{c^2}+\frac{c^2a^2}{b^2}\right)+2\left(a^2+b^2+c^2\right)\)

\(\ge\frac{1}{2}\cdot2\cdot\left(a^2+b^2+c^2\right)+2\cdot\left(a^2+b^2+c^2\right)\)

\(=3\cdot\left(a^2+b^2+c^2\right)=3\)

Do đó \(P\ge\sqrt{3}\)

Dấu đẳng thức xảy ra \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)

2. \(x^2+5x+9=\left(x+5\right)\sqrt{x^2+9}\)

Đặt \(\sqrt{x^2+9}=a>0\)

\(\Leftrightarrow a^2=x^2+9\)

\(pt\Leftrightarrow a^2+5x=\left(x+5\right)\cdot a\)

\(\Leftrightarrow a^2+5x-ax-5a=0\)

\(\Leftrightarrow a\left(a-x\right)-5\left(a-x\right)=0\)

\(\Leftrightarrow\left(a-x\right)\left(a-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=x\\a=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+9}=x\\\sqrt{x^2+9}=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+9=x^2\\x^2+9=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\varnothing\\x\in\left\{\pm4\right\}\end{matrix}\right.\)

Vậy...

Chữ hơi xấu thông cảm nha.

có thiếu gì ko bn

đầu bài k đầy đủ thì giải sao được

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

\(\Rightarrow\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-b-a}{c}-3=0\)

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

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

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

Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ne0\) nên chỉ có

x-a-b-c=0 =>x=a+b+c

Vậy x=a+b+c

ĐKXĐ : \(a+b\ne0;a+c\ne0;b+c\ne0.\)

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

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

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

\(\left(1\right)\) có vô số nghiệm \(\Leftrightarrow\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=0.\left(2\right)\)

Chẳng hạn ta chọn \(a=1,b=1.\)Để ( 2 ) xảy ra ta chọn c sao cho :

\(\frac{1}{2}+\frac{1}{1+c}+\frac{1}{1+c}=0\Leftrightarrow\frac{2}{1+c}=-\frac{1}{2}\Leftrightarrow c=-5.\)

Như vậy \(\left(1\right)\) có vô số nghiệm , chẳng hạn khi \(a=1,b=1,c=-5.\)

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