giải phương trình logarit

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nguyenbahiep1

Giải: [TEX]log_2(x^2+4x+6) + log_{\frac{1}{4}}(x^2+2) = 1 + log_8 (x+2)^3[/TEX]
Mình giải ra nghiệm lẻ quá không đoán nghiệm được

[laTEX]TXD: x > -2 \\ \\ log_2(x^2+4x+6) - log_2\sqrt{x^2+2} = log_22 + log_2(x+2) \\ \\ x^2+4x+6 = 2\sqrt{x^2+2}.(x+2) \\ \\ x = \pm \sqrt{2}, x = \frac{-4 \pm \sqrt{10}}{3}[/laTEX]
 
N

nguyenbahiep1

sau khi bình phương 2 vế

[laTEX]3x^4 +8x^3-4x^2-16x-4 = 0 \Rightarrow (x^2-2)(3x^2+8x+2) = 0 [/laTEX]
 
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