gpt ẩn x sau
\(\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}=a+b+c\)
K
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28 tháng 4 2015
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
KN
15 tháng 1 2020
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\)
\(\Leftrightarrow\frac{xa-a^2}{abc}+\frac{xb-b^2}{abc}+\frac{xc-c^2}{abc}=\frac{2bc}{abc}+\frac{2ac}{abc}+\frac{2ab}{abc}\)
\(\Leftrightarrow\frac{xa-a^2+xb-b^2+xc-c^2}{abc}=\frac{2bc+2ac+2ab}{abc}\)
\(\Leftrightarrow xa-a^2+xb-b^2+xc-c^2=2bc+2ac+2ab\)
\(\Leftrightarrow xa+xb+xc=2bc+2ac+2ab+a^2+b^2+c^2\)
\(\Leftrightarrow x\left(a+b+c\right)=\left(a+b+c\right)^2\)
\(\Leftrightarrow x=a+b+c\)
Vậy x = a + b + c
KN
15 tháng 1 2020
\(ĐKXĐ:a,b,c\ne0\)
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}=1-\frac{4x}{a+b+c}\)
\(\Leftrightarrow1+\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}=4\)
\(-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c\right)}{a+b+c}-\frac{4x}{a+b+c}\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}=\)
\(\frac{4\left(a+b+c-x\right)}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b+c}\right)=0\)
\(\Rightarrow\left(a+b+c-x\right)=0\)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
AH
Akai Haruma
Giáo viên
20 tháng 3 2020
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 $(1);(2)\Rightarrow x-a-b-c=0\Rightarrow x=a+b+c$
22 tháng 8 2019
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...
15 tháng 1 2017
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
1 tháng 1 2017
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 \(a=1,b=1,c=-5.\)
TK
2 tháng 1 2017
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pt \(\Rightarrow\frac{x}{a+b}+\frac{x}{a+c}+\frac{x}{b+c}=\left(\frac{ab}{a+b}+c\right)+\left(\frac{ac}{a+c}+b\right)+\left(\frac{bc}{b+c}+a\right)\)
\(\Rightarrow\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right).x=\frac{ab+ac+bc}{a+b}+\frac{ac+ab+bc}{a+c}+\frac{bc+ab+ac}{b+c}\)
\(\Rightarrow\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right).x=\left(ab+bc+ac\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\)
Nếu \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\ne0\) => phương trình có 1 ngiệm x = ab + bc +ca
Nếu \(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=0\) => phương trình có vô số nghiệm
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