\(\frac{5}{x}+\frac{4}{x+1}=\frac{3}{x+2}+\frac{2}{x+3}\)
ĐKXĐ: \(x\ne0;x\ne-1;x\ne-2;x\ne-3\)
\(\Leftrightarrow\left(\frac{5}{x}+1\right)+\left(\frac{4}{x+1}+1\right)=\left(\frac{3}{x+2}+1\right)+\left(\frac{2}{x+3}+1\right)\)
\(\Leftrightarrow\frac{5+x}{x}+\frac{4+x+1}{x+1}=\frac{3+x+2}{x+2}+\frac{2+x+3}{x+3}\)
\(\Leftrightarrow\frac{5+x}{x}+\frac{5+x}{x+1}=\frac{5+x}{x+2}+\frac{5+x}{x+3}\)
\(\Leftrightarrow\frac{5+x}{x}+\frac{5+x}{x+1}-\left(\frac{5+x}{x+2}+\frac{5+x}{x+3}\right)=0\)
\(\Leftrightarrow\left(5+x\right)\left(\left(\frac{1}{x}+\frac{1}{x+1}\right)-\left(\frac{1}{x+2}+\frac{1}{x+3}\right)\right)=0\) (*)
Vì \(\frac{1}{x}>\frac{1}{x+2};\frac{1}{x+1}>\frac{1}{x+3}\Rightarrow\frac{1}{x}+\frac{1}{x+1}>\frac{1}{x+2}+\frac{1}{x+3}\\ \Leftrightarrow\left(\frac{1}{x}+\frac{1}{x+1}\right)-\left(\frac{1}{x+2}+\frac{1}{x+3}\right)>0\)
Phương trình (*) xảy ra \(\Leftrightarrow5+x=0\Leftrightarrow x=-5\) (t/m ĐKXĐ)
Vậy phương trình có 1 nghiệm duy nhất là x = -5
\(\frac{5}{x}+\frac{4}{x+1}=\frac{3}{x+2}+\frac{2}{x+3}\\ \frac{5}{x}+1+\frac{4}{x+1}+1=\frac{3}{x+2}+1+\frac{2}{x+3}+1\\ \frac{5+x}{x}+\frac{5+x}{x+1}=\frac{5+x}{x+2}+\frac{5+x}{x+3}\\ \left(5+x\right).\left(\frac{1}{x}+\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\\ 5+x=0\\ x=-5\)
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