Cho phương trình:  \(
\frac{3ab+1}{a}x=\frac{3ab}{a+1}+\frac{\left (2a+1 \right )x}{a\left ( a+1 \right )^{2}}+\frac{a^{2}}{\left ( a+1 \right )^{3}}
\)(*)
a) Tìm điều kiện để phương trình có nghiệm.
b) Tính nghiệm $x$.
c) Tìm điều kiện để phương trình có nghiệm âm
a) Điều kiện cho a,b: \(
a\neq 0, a\neq -1.
\)
(*)\(
\Leftrightarrow [ \frac{3ab+1}{a}-\frac{2a+1}{a\left ( a+1\right )^{2}} ]x = \frac{3ab}{a+1}+\frac{a^{2}}{\left (a+1 \right )^{3}}
\)
\(
\Leftrightarrow [ \frac{a\left (3ab+1  \right )\left ( a+1\right )^{2}-\left ( 2a+1\right )}{a\left ( a+1\right )^{2}} ] x= \frac{3ab\left ( a+1\right )^{2}+a^{2}}{\left (a+1 \right )^{3}}
\)
\(
\Leftrightarrow \frac{\left (3ab+1  \right )\left ( a^{2}+2a+1 \right )-2a-1}{a\left ( a+1\right )^{2}}x=\frac{3ab\left ( a^{2}+2a+1 \right )+a^{2}}{\left (a+1 \right )^{3}}
\)
\(
\Leftrightarrow \frac{3a^{3}b+6a^{2}b+3ab+a^{2}}{a}x=\frac{3a^{3}b+6a^{2}b+3ab+a^{2}}{a+1}(**)
\)
Phương trình đã cho có nghiệm khi (**) có nghiệm
$\Leftrightarrow \left[\begin{array}{l}\frac{3a^{3}b+6a^{2}b+3ab+a^{2}}{a}\neq 0\\ \frac{3a^{3}b+6a^{2}b+3ab+a^{2}}{a}= \frac{3a^{3}b+6a^{2}b+3ab+a^{2}}{a}=0(L)  \end{array} \right.$
$\Leftrightarrow\left\{ \begin{array}{l}  3a^2b+6ab+3b+a\neq 0 \\ a\neq0\\a\neq-1 \end{array} \right. .$

b) Với điều kiện trên thì \(
3ab\left ( a+1\right )^{2}+a^{2}\neq 0 \Rightarrow \)Phương trình có nghiệm là \(x=\frac{a}{a+1}
\)

c) Để x<0 thì \(
\frac{a}{a+1} <0 \Leftrightarrow -1<a<0
\)

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