Question

4/2x+1-2/2x+3=1/2x+4-3/2x+2

Answer 1

ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};-1;\dfrac{-3}{2};-2\right\}\)

Ta có: \(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)

\(\Leftrightarrow\dfrac{4\left(2x+3\right)}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{2\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2}{\left(2x+2\right)\left(2x+4\right)}-\dfrac{3\left(2x+4\right)}{\left(2x+2\right)\left(2x+4\right)}\)

\(\Leftrightarrow\dfrac{8x+12-4x-2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{2x+2-6x-12}{\left(2x+2\right)\left(2x+4\right)}\)

\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}\)

\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}-\dfrac{-4x-10}{\left(2x+2\right)\left(2x+4\right)}=0\)

\(\Leftrightarrow\dfrac{4x+10}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{4x+10}{\left(2x+2\right)\left(2x+4\right)}=0\)

\(\Leftrightarrow\left(4x+10\right)\left(\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{1}{\left(2x+2\right)\left(2x+4\right)}\right)=0\)

\(\Leftrightarrow2\left(2x+5\right)\left(\dfrac{\left(2x+2\right)\left(2x+4\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}+\dfrac{\left(2x+1\right)\left(2x+3\right)}{\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)}\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(4x^2+8x+4x+8+4x^2+6x+2x+6\right)=0\)(Vì \(\left(2x+1\right)\left(2x+2\right)\left(2x+3\right)\left(2x+4\right)\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left(2x+5\right)\left(8x^2+20x+14\right)=0\)

mà \(8x^2+20x+14>0\forall x\)

nên 2x+5=0

\(\Leftrightarrow2x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{2}\)

Vậy: \(S=\left\{-\dfrac{5}{2}\right\}\)

Answer 2

Answer 3

\(\dfrac{4}{2x+1}-\dfrac{2}{2x+3}=\dfrac{1}{2x+4}-\dfrac{3}{2x+2}\)

\(\Leftrightarrow\dfrac{2x+5}{2x+1}-\dfrac{2x+5}{2x+3}=\dfrac{2x+5}{2x+4}-\dfrac{2x+5}{2x+2}\)

\(\Leftrightarrow\left(2x+5\right)\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+3}-\dfrac{1}{2x+4}+\dfrac{1}{2x+2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\\dfrac{1}{2x+1}-\dfrac{1}{2x+3}-\dfrac{1}{2x+4}+\dfrac{1}{2x+2}=0\left(1\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow\dfrac{2x+3-2x-1}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{2x+4-2x-2}{\left(2x+4\right)\left(2x+2\right)}=0\)

\(\Leftrightarrow\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}+\dfrac{2}{\left(2x+4\right)\left(2x+2\right)}=0\)

\(\Leftrightarrow\dfrac{1}{\left(2x+1\right)\left(2x+3\right)}=-\dfrac{1}{\left(2x+4\right)\left(2x+2\right)}\)

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(Vô lí)