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Решите задачу:

$\sqrt{1+log_2x}+\sqrt{4log_4x-2}}=4\\\\ODZ:\; \; \left \{ {{1+log_2x\geq 0} \atop {4log_4x-2\geq 0}} \right. \; \left \{ {{log_2x\geq -1} \atop {log_4x\geq \frac{1}{2}}} \right. \; \left \{ {{x\geq \frac{1}{2}} \atop {x\geq 2}} \right. \; \; \to \; \; x\geq 2\\\\4log_4x=4log_{2^2}x=4\cdot \frac{1}{2}log_2x=2log_2x\\\\t=log_2x\; ,\; \; \sqrt{1+t}+\sqrt{2t-2}=4\\\\\sqrt{1+t}=4-\sqrt{2t-2}\\\\1+t=16-8\sqrt{2t-2}+2t-2\\\\8\sqrt{2t-2}=t+13\\\\64(2t-2)=t^2+26t+169\\\\t^2-102t+297=0\\\\D/4=51^2-297=2304,\; \; \sqrt{D}=48$

$t_1=51-48=3\; ,\; \; t_2=51+48=99\\\\log_2x=3\; \; \to \; \; x=2^3=8\in ODZ\\\\log_2x=99\; ,\; \; x=2^{99}$

$Proverka:x=2^{99},\sqrt{1+log_22^{99}}+\sqrt{4log_42^{99}-2}=4,\\\\\sqrt{1+99}+\sqrt{2*99-2}=4\\\\10+13=4\; \; neverno\\\\x=2^3,\sqrt{1+log_22^3}+\sqrt{4log_42^3-2}=4\\\\\sqrt{1+3}+\sqrt{6-2}=4\\\\2+2=4\\\\4=4\; \; verno\\\\Otvet:x=8$