设不等式2(log1/2^x)^2+9(log1/2^x)+9≤0的解集为M,求M
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设不等式2(log1/2^x)^2+9(log1/2^x)+9≤0的解集为M,求M
设不等式2(log1/2^x)^2+9(log1/2^x)+9≤0的解集为M,求M
设不等式2(log1/2^x)^2+9(log1/2^x)+9≤0的解集为M,求M
令log‹1/2›x=u,则原式可化为2u²+9u+9=(2u+3)(u+3)=2(u+3/2)(u+3)≦0,于是得:
-3≦u≦-3/2,即有-3≦log‹1/2›x≦-3/2;
由log‹1/2›x≧-3,得0
LOG底数是几啦,不知道底数怎么帮你解答啦
设不等式2log²‹1/2›x+9log‹1/2›x+9≤0的解集为M,求M
令log‹1/2›x=u,则原式可化为2u²+9u+9=(2u+3)(u+3)=2(u+3/2)(u+3)≦0,于是得:
-3≦u≦-3/2,即有-3≦log‹1/2›x≦-3/2;
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设不等式2log²‹1/2›x+9log‹1/2›x+9≤0的解集为M,求M
令log‹1/2›x=u,则原式可化为2u²+9u+9=(2u+3)(u+3)=2(u+3/2)(u+3)≦0,于是得:
-3≦u≦-3/2,即有-3≦log‹1/2›x≦-3/2;
由log‹1/2›x≧-3,得0
①∩②={x∣4^(1/3)≦x≦8}=M.
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