a)(4x/x^2+4x+3)-1=6.((1/x+3)-(1/2x+2))
b)(-1/1-x)+(2x^2-5/x^3-1)=(4/x^2+x+1)
c)(12x+1/6x-2)-(9x-5/3x-1)=(108x-36x^2-9/4(9x^2-1)
a)(4x/x^2+4x+3)-1=6.((1/x+3)-(1/2x+2))
b)(-1/1-x)+(2x^2-5/x^3-1)=(4/x^2+x+1)
c)(12x+1/6x-2)-(9x-5/3x-1)=(108x-36x^2-9/4(9x^2-1)
Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne – 1;x \ne – 3\\
\dfrac{{4x}}{{{x^2} + 4x + 3}} – 1 = 6.\left( {\dfrac{1}{{x + 3}} – \dfrac{1}{{2x + 2}}} \right)\\
\Rightarrow \dfrac{{4x – {x^2} – 4x – 3}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \dfrac{6}{{x + 3}} – \dfrac{3}{{x + 1}}\\
\Rightarrow \dfrac{{ – {x^2} – 3}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \dfrac{{6\left( {x + 1} \right) – 3\left( {x + 3} \right)}}{{\left( {x + 1} \right)\left( {x + 3} \right)}}\\
\Rightarrow – {x^2} – 3 = 6x + 6 – 3x – 3\\
\Rightarrow {x^2} + 3x + 6 = 0\left( {vn} \right)\\
\text{Vậy pt vô nghiệm}\\
b)Dkxd:x \ne 1\\
\dfrac{{ – 1}}{{1 – x}} + \dfrac{{2{x^2} – 5}}{{{x^3} – 1}} = \dfrac{4}{{{x^2} + x + 1}}\\
\Rightarrow \dfrac{1}{{x – 1}} + \dfrac{{2{x^2} – 5}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{4}{{{x^2} + x + 1}}\\
\Rightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} – 5}}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{4\left( {x – 1} \right)}}{{{x^2} + x + 1}}\\
\Rightarrow 3{x^2} + x – 4 = 4x – 4\\
\Rightarrow 3{x^2} – 3x = 0\\
\Rightarrow 3x\left( {x – 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\left( {ktm} \right)\\
\text{Vậy}\,x = 0\\
c)Dkxd:x \ne \dfrac{1}{3};x \ne – \dfrac{1}{3}\\
\dfrac{{12x + 1}}{{6x – 2}} – \dfrac{{9x – 5}}{{3x – 1}} = \dfrac{{108x – 36{x^2} – 9}}{{4\left( {9{x^2} – 1} \right)}}\\
\Rightarrow \dfrac{{12x + 1 – 18x + 10}}{{2\left( {3x – 1} \right)}} = \dfrac{{108x – 36{x^2} – 9}}{{4\left( {3x – 1} \right)\left( {3x + 1} \right)}}\\
\Rightarrow \dfrac{{\left( {11 – 6x} \right).2.\left( {3x + 1} \right)}}{{4\left( {3x – 1} \right)\left( {3x + 1} \right)}} = \dfrac{{108x – 36{x^2} – 9}}{{\left( {3x – 1} \right)\left( {3x + 1} \right)}}\\
\Rightarrow – 36{x^2} + 66x – 12x + 22 = 108x – 36{x^2} – 9\\
\Rightarrow 108x – 66x + 12x = 22 + 9\\
\Rightarrow 54x = 31\\
\Rightarrow x = \dfrac{{31}}{{54}}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{{31}}{{54}}
\end{array}$