Toán Lớp 8: Tính `(5x-2)/((x-2)(x+2))-3/(x+2)+x/(x-2)`

Question

Toán Lớp 8:  Tính (5x-2)/((x-2)(x+2))-3/(x+2)+x/(x-2)

lyly98 · Answer

{5x-2}/{(x-2)(x+2)}- 3/{x+2}+ x/{x-2}   = {(5x-2)-[3(x-2)]+[x(x+2)]}/{(x-2)(x+2)}   = {5x-2-3x+6+x^2+2x}/{(x-2)(x+2)}   = {(x^2+4x+4)}/{(x-2)(x+2)}   = {(x+2)^2}/{(x-2)(x+2)}   = {x+2}/{x-2}

thuminhthong · Answer

#andy     [ begin{array}{l}  dfrac{{5x - 2}}{{ left( {x - 2}  right) left( {x + 2}  right)}} -  dfrac{3}{{x + 2}} +  dfrac{x}{{x - 2}}    =  dfrac{{5x - 2}}{{ left( {x - 2}  right) left( {x + 2}  right)}} -  dfrac{{3 left( {x - 2}  right)}}{{ left( {x - 2}  right) left( {x + 2}  right)}} +  dfrac{{x left( {x + 2}  right)}}{{ left( {x - 2}  right) left( {x + 2}  right)}}    =  dfrac{{5x - 2 - 3x + 6 + {x^2} + 2x}}{{ left( {x - 2}  right) left( {x + 2}  right)}}    =  dfrac{{{x^2} + 4x + 4}}{{ left( {x - 2}  right) left( {x + 2}  right)}}    =  dfrac{{{{ left( {x + 2}  right)}^2}}}{{ left( {x - 2}  right) left( {x + 2}  right)}}    =  dfrac{{x + 2}}{{x - 2}}  end{array} ]