[Toán 11] Giải PT

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nguyenbahiep1

[TEX](4x-1).\sqrt[3]{2-8{x}^{3}}=2x[/TEX]
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[laTEX] u = \sqrt[3]{2-8{x}^{3}} \Rightarrow u^3 = 2- (2x)^3 \\ \\ u^3+(2x)^3 = 2 \\ \\ (4x-1).u= 2x \\ \\ 2x = v \\ \\ u^3 +v^3 = 2 \Rightarrow(u+v).[(u+v)^2-3uv] = 2 \\ \\ (2v-1).u = v \Rightarrow 2u.v = u+v \\ \\ 2u.v. [ 4(u.v)^2 -3uv] = 2 \\ \\ 4(uv)^3 - 3(uv)^2 - 1= 0 \Rightarrow uv = 1 \Rightarrow u+v = 2 \\ \\ u = v = 1 \Rightarrow 2x = 1\Rightarrow x = \frac{1}{2}[/laTEX]
 
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