Toán Lớp 8: X-1/x-2 + x+3/x-4 = 2/(x-2)(x-4)

Question

Toán Lớp 8:  X-1/x-2 + x+3/x-4 = 2/(x-2)(x-4)

kimnguenoanh · Answer

Giải đáp:    S ={1+ sqrt{3};1- sqrt{3}}    Lời giải và giải thích chi tiết:    (x-1)/(x-2) + (x+3)/(x-4) = 2/((x-2)(x-4)) (dkxđ: $x neq2 ; x neq4)$     &hArr;((x-1)(x-4))/((x-4)(x-2)) + ((x+3)(x-2))/((x-2)(x-4))= 2/((x-2)(x-4))     &rArr;(x-1)(x-4) + (x+3)(x-2)=2     &hArr;x^2-4x-x+4+x^2-2x+3x-6-2=0     &hArr;2x^2-4x - 4 =0     &hArr;2(x^2-2x-2)=0      &hArr;x^2-2x-2 =0     &hArr;x^2-2x+1 =3     &hArr;(x-1)^2=3     &hArr;  ( left[  begin{array}{l}x=1+ sqrt{3}(tm)  x=1- sqrt{3}(tm) end{array}  right. )      Vậy S ={1+ sqrt{3};1- sqrt{3}}

cobebongdem · Answer

Giải đáp:$(x neq2;4)$     $ dfrac{x-1}{x-2}+ dfrac{x+3}{x-4}= dfrac{2}{(x-2)(x-4)}$     $&hArr; dfrac{(x-1)(x-4)}{(x-2)(x-4)}+ dfrac{(x+3)(x-2)}{(x-4)(x-2)}- dfrac{2}{(x-2)(x-4)}=0$     $&hArr; dfrac{x^2-5x+4+x^2+x-6-2}{(x-4)(x-2)}=0$     $&hArr;2x^2-4x-4=0$     $&hArr;x^2-2x-2=0$     $&hArr;x^2-2x+1=3$     $&hArr;(x-1)^2=3$     $&hArr; left[  begin{array}{l}x= sqrt{3}+1(t/m)   x=- sqrt{3}+1(t/m) end{array}  right.$