(x-2)4 + (x-3)4 = 1
(8x-7)2(4x+3)(x+1)= 3,5
x3 - 6x2 + 4x + 6 = 0
2(x-2)(3-2x)2 + (2x-1)3 + 52(x-1)(x+3) = 0
(x-2)4 + (x-3)4 = 1
(8x-7)2(4x+3)(x+1)= 3,5
x3 - 6x2 + 4x + 6 = 0
2(x-2)(3-2x)2 + (2x-1)3 + 52(x-1)(x+3) = 0
a) Đặt $t=x-\frac{5}{2}$. Khi đó pt trở thành:
$(t+\frac{1}{2})^{4}+(t-\frac{1}{2})^{4}=1\Leftrightarrow 2t^{4}+3t^{2}-\frac{7}{8}=0\Leftrightarrow t^{2}=\frac{1}{4}\Leftrightarrow t=\frac{1}{2}ort=\frac{-1}{2}\Leftrightarrow x=3orx= 2$
"Life would be tragic if it weren't funny"
-Stephen Hawking-
0 thành viên, 1 khách, 0 thành viên ẩn danh