c: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3=4+3x^2+9x-x-3\)
\(\Leftrightarrow3x^2+5x-8=3x^2+8x+1\)
=>8x+1=5x-8
=>3x=-9
=>x=-3(loại)
d: \(\Leftrightarrow\dfrac{-1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-...-\dfrac{1}{x-5}+\dfrac{1}{x-6}=\dfrac{1}{8}\)
=>-1/x-2+1/x-6=1/8
=>\(\dfrac{-x+6+x-2}{\left(x-2\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>4/(x-2)(x+6)=1/8
=>x^2+4x-12=32
=>x^2+4x-44=0
=>\(x=-2\pm4\sqrt{3}\)
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. Ta có:
