高一数学 9^(-x)-2*3^(1-x)=27

\u9ad8\u4e00\u6570\u5b66\u9898\uff1a\u89e3\u65b9\u7a0b6^x\uff0b4^x\uff1d9^x\uff0c9^\uff08-x\uff09-2*3^\uff081-x\uff09=27

6^x\uff0b4^x\uff1d9^x
3^x*2^x+(2^x)^2=(3^x)^2,
\u4e24\u8fb9\u540c\u9664\u4ee5(2^x)^2\uff0c
\uff083/2\uff09^x+1=[(3/2)^x]^2,
\u8bbe\uff083/2\uff09^x=t,
t+1=t^2,
t^2-t-1=0,
t=(1+\u221a5\uff09/2\uff0c\uff08\u820d\u53bb\u8d1f\u6839\uff09\uff0c
\uff083/2\uff09^2=(1+\u221a5\uff09/2,\u4e24\u8fb9\u53d6\u5bf9\u6570\uff0c
x=[lg(1+\u221a5\uff09-lg2]/(lg3-lg2).




9^\uff08-x\uff09-2*3^\uff081-x\uff09=27
1/[\uff083^2)^x]-6/(3^x)=27,
\u8bbe1/3^x=t,
t^2-6t-27=0
(t-9)(t+3)=0
t=9,\u6216t=-3(\u4e0d\u7b26\u5408\u9898\u610f\uff0c\u820d\u53bb\uff09
1/3^x=9,
3^(-x)=3^2
-x=2
x=-2
\u7ecf\u68c0\u9a8cx=-2\u662f\u539f\u65b9\u7a0b\u7684\u89e3\u3002

1/81

9^(-x)-2*3^(1-x)=27
(3*3)^(-x) - 2 * 3 * 3^(-x) = 27
3^(-x) * 3^(-x) - 6 * 3^(-x) - 27 = 0
[3^(-x) - 9] [3^(-x) + 3] = 0
3^(-x) = 9
3^(-x) = -3

3^(-x) 始终大于0, 所以 3^(-x) = -3 无解

所以
3^(-x) = 9
x = -2

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