a:
ĐKXĐ: \(x\notin\left\{-3;-\dfrac{1}{3}\right\}\)
\(A=\dfrac{3x-1}{3x+1}+\dfrac{x-3}{x+3}\)
\(=\dfrac{\left(3x-1\right)\left(x+3\right)+\left(x-3\right)\left(3x+1\right)}{\left(x+3\right)\left(3x+1\right)}\)
\(=\dfrac{3x^2+9x-x-3+3x^2+x-9x-3}{\left(x+3\right)\left(3x+1\right)}=\dfrac{6x^2-6}{\left(x+3\right)\left(3x+1\right)}\)
Để A=2 thì \(6x^2-6=2\left(x+3\right)\left(3x+1\right)\)
=>\(6x^2-6=\left(2x+6\right)\left(3x+1\right)\)
=>\(6x^2+2x+18x+6=6x^2-6\)
=>20x=-12
=>\(x=-\dfrac{12}{20}=-\dfrac{3}{5}\left(nhận\right)\)
b:
ĐKXĐ: x<>-3
\(B=\dfrac{10}{3}-\dfrac{3x-1}{4x+12}-\dfrac{7x+2}{6x+18}\)
\(=\dfrac{10}{3}-\dfrac{3x-1}{4\left(x+3\right)}-\dfrac{7x+2}{6\left(x+3\right)}\)
\(=\dfrac{10\cdot4\left(x+3\right)}{12\left(x+3\right)}-\dfrac{3\left(3x-12\right)}{12\left(x+3\right)}-\dfrac{2\left(7x+2\right)}{12\left(x+3\right)}\)
\(=\dfrac{40x+12-9x+36-14x-4}{12\left(x+3\right)}\)
\(=\dfrac{17x+44}{12x+36}\)
Để B=2 thì 17x+44=2(12x+36)
=>24x+72=17x+44
=>7x=44-72=-28
=>x=-4(nhận)
help me!!!!!
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