\(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=-2,5\)
\(\Rightarrow\frac{4x+1}{\left(x+1\right)\left(x-2\right)}+\frac{2,5\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Rightarrow4x+1+2,5x^2-5x+2,5x-5=0\)
\(\Rightarrow2,5x^2+1,5x-4=0\)
\(\Rightarrow2,5x^2-2,5x+4x-4=0\)
\(\Rightarrow2,5x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2,5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2,5x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1,6\end{matrix}\right.\)
Vậy....
Ta có: \(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=-2,5\)
⇔\(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=\frac{-5}{2}\)
⇔\(2\left(4x+1\right)=-5\left(x+1\right)\left(x-2\right)\)
⇔\(8x+2=\left(-5x-5\right)\left(x-2\right)\)
⇔\(8x+2=-5x^2+5x+10\)
⇔\(8x+2+5x^2-5x-10=0\)
⇔\(3x+5x^2-8=0\)
⇔\(5x^2+3x-8=0\)
⇔\(5x^2-5x+8x-8=0\)
⇔\(5x\left(x-1\right)+8\left(x-1\right)=0\)
⇔\(\left(x-1\right)\left(5x+8\right)=0\)
⇔\(\left[{}\begin{matrix}x-1=0\\5x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-8}{5}=-1,6\end{matrix}\right.\)
Vậy: x∈{1;-1,6}
\(\frac{4x+1}{\left(x+1\right)\left(x-2\right)}=-2.5\) \(\left(ĐKXĐ:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow-10\left(x+1\right)\left(x-2\right)=4x+1\)
\(\Leftrightarrow10x^2-10x-2=4x+1\)
\(\Leftrightarrow10x^2-14x-3=0\)
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