Question

a) \(\dfrac{5x-2}{2-2x}\)+\(\dfrac{2x-1}{2}\)=1-\(\dfrac{x^2+x-3}{1-x}\)

b)\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}\)=\(\dfrac{x\left(3x-1\right)+1}{\left(x-2\right).\left(x-2\right)}\)

c)1+\(\dfrac{x}{3-x}\)=\(\dfrac{3x}{\left(x+2\right).\left(x-3\right)}+\dfrac{2}{x+2}\)

Answer 1

a,\(\dfrac{5x-2}{2-2x}+\dfrac{2x-1}{2}=1-\dfrac{x^2-x-3}{1-x}\)

<=>\(\dfrac{5x-2}{2\left(1-x\right)}+\dfrac{2x-1}{2}=1-\dfrac{x^2-x-3}{1-x}\)

<=>\(\dfrac{5x-2}{2\left(1-x\right)}+\dfrac{\left(2x-1\right)\left(1-x\right)}{2\left(1-x\right)}=\dfrac{2\left(1-x\right)}{2\left(1-x\right)}-\dfrac{2\left(x^2-x-3\right)}{2\left(1-x\right)}\)

=>\(5x-2+2x-2x^2-1+x=2-2x-2x^2+2x+6\)

<=>\(-2x^2+8x-3=-2x^2+8\)

<=>\(8x=11< =>x=\dfrac{11}{8}\)

vậy..........

b,\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-1\right)+1}{\left(x-2\right)\left(x+2\right)}\)

<=>\(\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(3x-1\right)+1}{\left(x-2\right)\left(x+2\right)}\)

=>\(x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-x+1\)

<=>\(3x^2-25x-6=3x^2-x+1\)

<=>\(-24x=7< =>x=\dfrac{-7}{24}\)

vậy..................

câu c tương tự nhé :)