x^2+2x+5=0
=>x^2+2x+1+4=0
=>(x+1)^2=-4(vô lý)
=>\(S_1=\varnothing\left(1\right)\)
\(\dfrac{x-2}{3x+1}-\dfrac{x}{x-2}=1+\dfrac{7}{\left(3x+1\right)\left(x-2\right)}\)
=>\(\dfrac{x^2-4x+4-3x^2-x}{\left(3x+1\right)\left(x-2\right)}=\dfrac{3x^2-6x+x-2+7}{\left(3x+1\right)\left(x-2\right)}\)
=>3x^2-5x+5=-2x^2-5x+4
=>5x^2=-1(vô lý)
=>\(S_2=\varnothing\left(2\right)\)
(1),(2) suy ra hai phương trình tương đương
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