Toán Lớp 8: $A=(\frac{x+3}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}):(1-\frac{x-1}{x+3}$
Tìm x để $A=\frac{1}{2}$
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Toán Lớp 8: $A=(\frac{x+3}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}):(1-\frac{x-1}{x+3}$
Tìm x để $A=\frac{1}{2}$
Comments ( 1 )
ĐKXĐ : x\ne ±3
A=((x+3)/(x-3)+18/(9-x^2) +(x-3)/(x+3)) : (1-(x-1)/(x+3))
= (x^2+6x+9 – 18 + x^2-6x+9)/((x-3)(x+3)) : (x+3 -x+1)/(x+3)
= (2x^2)/((x-3)(x+3)) . (x+3)/4
= x^2/(2(x-3))
Để A=1/2
->x^2/(2x-6)=1/2
-> 2x^2=2x-6
->2x^2-2x+6=0
->x^2-x+3=0
-> x^2-2.x. 1/2+(1/2)^2 +11/4=0
->(x-1/2)^2+11/4=0
Do (x-1/2)^2+11/4>= 11/4>0
-> không có x thỏa mãn
Vậy không có x thỏa mãn để A=1/2