giải giúp em pt logarit

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nguyenbahiep1

[TEX]\frac32log_1/4(x+2)^2-3=log_1/4 (4-x)^3+log_1/4(x+6)^3[/TEX]
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[laTEX]\frac{3}{2}log_{\frac{1}{4}}(x+2)^2 - 3 = log_{\frac{1}{4}}(4-x)^3 + log_{\frac{1}{4}}(x+6)^3 \\ \\ txd: -6 < x < 4 , n \not = - 2 \\ \\ 3log_{\frac{1}{4}}|x+2| - 3 = 3log_{\frac{1}{4}}(4-x) + 3log_{\frac{1}{4}}(x+6) \\ \\ 4|x+2| = (4-x)(x+6) \\ \\ TH_1: -2 < x < 4 \\ \\ \Rightarrow 4(x+2) = (4-x)(x+6) \\ \\ TH_2: -6 < x < -2 \Rightarrow -4(x+2) = (4-x)(x+6)[/laTEX]
 
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