how do i solve x/x^2-4 + 1/x+2=3

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how do i solve x/x^2-4 + 1/x+2=3

Mathematics
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x/x^2-4 + 1/x+2=3 x/(x-2)(x+2) + 1/x+2=3 x+x-2/(x-2)(x+2)=3 2x-2=3 *(x-2)(x+2) 2x-2=3(x^2 - 4) 3x^2 -2x-2=0 3x^2 -3x + x -2=0 3x(x -1) + 2(x -1)=0 (3x+2)((x - 1) =0 x= -2/3, 1 Hope I did not make any mistake ..let me know if thats correct or not?
oh bad luck...found a mistake...thats the drawback of now writing but typing it
let me correct it again

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Other answers:

3x^2 -2x-10=0 x= 2 +- sqrt (4+120) /6 x= 2+- 2 sqrt(31) /6 x= 1+- sqrt (31)/ 3 hope this time its correct...pls reply to let me know
what is sqrt?
square root
well im completely confused because i dont know how to solve the problem..you lost me at the third step
i took the lcm of (x-2)(x+2) and (x+2) which is (x-2)(x+2
and shouldnt you have distributed the lcm throughout both sides?
the lcm which we got , was multiplied with numerator on the right hand side which was 3
fan me pls if you liked my help
ok thanks..
fan me plsssssssssss
tnx ...
welcum..
gud luck with ur maths...

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