Решите неравенство (log22x−log2x)2+(\log_2^2 x-\log_2x)^2+(log22x−log2x)2+32log2x+48<20log22x32\log_2x+48 < 20\log_2^2 x32log2x+48<20log22x▸Ответx∈(116;12)∪(8;16)x\in\left(\dfrac1{16} ; \dfrac12 \right)\cup(8;16)x∈(161;21)∪(8;16)▸Решение(log22x−log2x)2+(\log_2^2 x-\log_2x)^2+(log22x−log2x)2+32log2x+48<20log22x32\log_2x+48 < 20\log_2^2 x32log2x+48<20log22x ОДЗ: x>0x>0x>0 Пусть log2x=t\log_2x=tlog2x=t (t2−t)2+(t^2-t)^2+(t2−t)2+32t+48<20t232t+48 < 20t^232t+48<20t2 t4−2t3+t2+t^4-2t^3+t^2+t4−2t3+t2+32t+48−20t2<032t+48-20t^2 < 032t+48−20t2<0 t4−2t3−19t2+32t+48<0t^4-2t^3-19t^2+32t+48 < 0t4−2t3−19t2+32t+48<0 −2t(t2−16)+t4−19t2+48<0-2t(t^2-16)+t^4-19t^2+48 < 0−2t(t2−16)+t4−19t2+48<0 Попробуем разбить на множители t4−19t2+48t^4-19t^2+48t4−19t2+48. Для этого решим биквадратное уравнение. Введем замену:t2=yt^2=yt2=y y2−19y+48=0y^2-19y+48=0y2−19y+48=0 D=(−19)2−4⋅48=D=(-19)^2-4\cdot 48=D=(−19)2−4⋅48=361−192=169361-192=169361−192=169 y1=−(−19)+1692=y_1=\frac{-(-19)+\sqrt{169} }{2}=y1=2−(−19)+169=19+132=16\dfrac{19+13}2=16219+13=16 y2=19−132=62=3y_2=\dfrac{19-13}2=\dfrac62=3y2=219−13=26=3 t4−19t2+48=(t2−16)(t2−3)t^4-19t^2+48=(t^2-16)(t^2-3)t4−19t2+48=(t2−16)(t2−3) −2t(t2−16)+-2t(t^2-16)+−2t(t2−16)+(t2−16)(t2−3)<0(t^2-16)(t^2-3) < 0(t2−16)(t2−3)<0 (t2−16)(t2−2t−3)<0(t^2-16)(t^2-2t-3)< 0(t2−16)(t2−2t−3)<0 t2−2t−3=0t^2-2t-3=0t2−2t−3=0 D=(−2)2−4⋅(−3)=4+12=D=(-2)^2-4\cdot (-3)=4+12=D=(−2)2−4⋅(−3)=4+12=161616 t1=−(−2)+162=2+42=3t_1=\frac{-(-2)+\sqrt{16} }{2}=\dfrac{2+4}2=3t1=2−(−2)+16=22+4=3 t2=2−42=−22=−1t_2=\dfrac{2-4}2=-\dfrac22=-1t2=22−4=−22=−1 (t−4)(t+4)(t−3)(t+1)<0(t-4)(t+4)(t-3)(t+1) < 0(t−4)(t+4)(t−3)(t+1)<0 1)−4<t<−11)-4 < t< -11)−4<t<−1 −4<log2x<−1-4 < \log_2x < -1−4<log2x<−1 116<x<12\dfrac1{16} < x < \dfrac12161<x<21 2)3<t<42) 3< t< 42)3<t<4 3<log2x<43 < \log_2x < 43<log2x<4 8<x<168 < x < 168<x<16
