Toán Lớp 9: căn bậc hai(x-1)+căn bậc hai(x+3)+2*căn bậc hai((x-1)*(x^2 -3*x+5)) = 4 -2*x
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Toán Lớp 9: căn bậc hai(x-1)+căn bậc hai(x+3)+2*căn bậc hai((x-1)*(x^2 -3*x+5)) = 4 -2*x
Comments ( 1 )
Nếu $ x > 1 :$
$ VT > \sqrt{1 + x} > \sqrt{1 + 3} = 2$
$ VP = 4 – 2x < 4 – 2 = 2$
Vậy Chỉ có $ x = 1 => VT = VP = 2 (TM)$