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Solve-x-2-3x-x-3-4x-2-x-2-2x-1-x-2-

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Question Number 217262 by ArshadS last updated on 07/Mar/25
Solve: ((x^2 +3x)/(x^3 −4x))−(2/(x^2 +2x))=(1/(x−2))
$${Solve}: \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}}=\frac{\mathrm{1}}{{x}−\mathrm{2}} \\ $$
Answered by ibrahimmatematic last updated on 07/Mar/25
((x(x+3))/(x(x−2)(x+2)))−(2/(x(x+2)))=(1/(x−2)) ((x(x+3))/(x(x−2)(x+2)))=(1/(x−2))+(2/(x(x+2))) ((x(x+3))/(x(x+2)(x−2)))=((x(x+2)+2(x−2))/(x(x−2)(x+2))) x^2 +3x=x^2 +2x+2x−4 4x−4=3x x=4
$$\frac{{x}\left({x}+\mathrm{3}\right)}{{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}−\frac{\mathrm{2}}{{x}\left({x}+\mathrm{2}\right)}=\frac{\mathrm{1}}{{x}−\mathrm{2}} \\ $$$$\frac{{x}\left({x}+\mathrm{3}\right)}{{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}=\frac{\mathrm{1}}{{x}−\mathrm{2}}+\frac{\mathrm{2}}{{x}\left({x}+\mathrm{2}\right)} \\ $$$$\frac{{x}\left({x}+\mathrm{3}\right)}{{x}\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}=\frac{{x}\left({x}+\mathrm{2}\right)+\mathrm{2}\left({x}−\mathrm{2}\right)}{{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$${x}^{\mathrm{2}} +\mathrm{3}{x}={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}{x}−\mathrm{4} \\ $$$$\mathrm{4}{x}−\mathrm{4}=\mathrm{3}{x} \\ $$$${x}=\mathrm{4} \\ $$
Answered by Rasheed.Sindhi last updated on 07/Mar/25
((x(x+3))/(x(x−2)(x+2)))−(2/(x(x+2)))=(1/(x−2)) ;x≠0,±2 (5/(4(x−2)))−(1/(4(x+2)))−(1/x)+(1/(x+2))=(1/(x−2)) (5/(4(x−2)))−(1/(x−2))−(1/(4(x+2)))+(1/(x+2))−(1/x)=0 ((5−4)/(4(x−2)))+((−1+4)/(4(x+2)))−(1/x)=0 (1/(4(x−2)))+(3/(4(x+2)))=(1/x) (1/(x−2))+(3/(x+2))=(4/x) ((x+2+3x−6)/(x^2 −4))=(4/x) ((4x−4)/(x^2 −4))=(4/x) ((x−1)/(x^2 −4))=(1/x) x^2 −x=x^2 −4 x=4
$$\frac{\cancel{{x}}\left({x}+\mathrm{3}\right)}{\cancel{{x}}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)}−\frac{\mathrm{2}}{{x}\left({x}+\mathrm{2}\right)}=\frac{\mathrm{1}}{{x}−\mathrm{2}}\:;{x}\neq\mathrm{0},\pm\mathrm{2} \\ $$$$\frac{\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{2}\right)}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}+\mathrm{2}}=\frac{\mathrm{1}}{{x}−\mathrm{2}} \\ $$$$\frac{\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{2}\right)}−\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}\left({x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{{x}+\mathrm{2}}−\frac{\mathrm{1}}{{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{5}−\mathrm{4}}{\mathrm{4}\left({x}−\mathrm{2}\right)}+\frac{−\mathrm{1}+\mathrm{4}}{\mathrm{4}\left({x}+\mathrm{2}\right)}−\frac{\mathrm{1}}{{x}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}\left({x}−\mathrm{2}\right)}+\frac{\mathrm{3}}{\mathrm{4}\left({x}+\mathrm{2}\right)}=\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{2}}+\frac{\mathrm{3}}{{x}+\mathrm{2}}=\frac{\mathrm{4}}{{x}} \\ $$$$\frac{{x}+\mathrm{2}+\mathrm{3}{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{4}}{{x}} \\ $$$$\frac{\mathrm{4}{x}−\mathrm{4}}{{x}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{4}}{{x}} \\ $$$$\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} −{x}={x}^{\mathrm{2}} −\mathrm{4} \\ $$$${x}=\mathrm{4} \\ $$

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