方程4的x次方 6的x次方=6.9的x次方-搜答案-搜搜考试网

4^x+6^x=9^x(2^x)^2+2^x*3^x-(3^x)^2=0把2^x看成未知数来求,得2^x=(根号5-1)*3^x/2整理一下：（3/2)^x=2/(根号5-1）所以x=log3/2为底

两边都乘以X^2-4x=5

x^5-3x^4+2x^3-3x^2+9x-6=0X^3(x^2-3x+2)-3(x^2-3x+2)=0(x^2-3x+2)(X^3-3)=0(x-1)(x-2)(X^3-3)=0x=1或2或3的根号

t=3^x>0t^2-6t-7=0(t-7)(t+1)=0t=7x=log3(7)

4^x+6^x=9^x(4/6)^x+1=(9/6)^x(2/3)^x+1=(3/2)^x((2/3)^x)^2+(2/3)^x-1=0设(2/3)^x=a则：a^2+a-1=0,a>0a=(√5-1

设a=2^x,b=3^x那么方程是：3a^2+2b^2=5ab(3a-2b)(a-b)=0(1):3a-2b=03*2^x=2*3^x(2/3)^x=2/3x=1(2)a-b=03^x=2^x(3/2

设3^x=M,2^x=N,则9^x=3^2x=（3^x）^2=M^2,6^x=MN,2^（2x+1）=2N^2,于是原方程化为：M^2+MN-2N^2=0,分解因式得：（M-2N)(M+N)=0,故M

(x平方+3)(x平方-2)=0(x平方+3)(x+√2)(x-√2)=0∴x=-√2x=√2

设x²=t则t＞0t²-t-6=0(t-3)(t+2)=0t=3或-2（舍）∴x=±√3

9^x+4^x=5/2*6^x两边同时除以6^x得：（3/2)^x+(2/3)^x=5/2令t=(3/2)^x,则方程化为;t+1/t=5/2t^2-5/2*t+1=0(t-2)(t-1/2)=0t=

约为0.93,帮你弄了两个过程,你看下吧1)设6^x=a4^x=b9^x=a+b那么log(6)a=log(4)b=log(9)a+b用换底公式lga/lg6=lgb/lg4=lga+b/lg9然后整

4^x+6^x=9^x方程两边同除以9^x得：(4/9)^x+(2/3)^x=1令t=(2/3)^x则,(4/9)^x=t²原式为；t²+t-1=0由求根公式得；t=(-1±√5)

6^x+4^x=9^x两边同除以9^x,可得：(2/3)^x+(2/3)^(2x)=1,令y=(2/3)^x,可得：y+y^2=1,即：y^2+y-1=0,解得：y=(√5-1)/2,（舍去负值）即：

4^x+4^(-x)-6*[2^x+2^(-x)]+10=0设2^x+2^(-x)=t＞0t^2-6t+8=0(t-2)(t-4)=0t=2或t=42^x+2^(-x)=2x=02^x+2^(-x)=

令3^x=a则9^x=a²a²-6x-7=0(a-7)(a+1)=0a=3^x>0所以3^x=a=7x=log3(7)

1/2的X次方乘以8的2X次方=42的-X次方乘以2的6X次方=2的2次方6x-x=25x=2x=0.4再问：化简2的4+log23的次方

∵x的6次方>=0∴x的6次方+2>0∴方程x的6次方+2=0无解

-2答案不唯一

(x+1)(x+4)=x的2次方-6x²+5x+4=x²-65x=-10x=-2

是4^x+6^x=6*9^x吧?两边同除以9^x:(4/9)^x+(6/9)^x=6即(2/3)^2x+(2/3)^2-6=0[(2/3)^x+3][(2/3)^x-2]=0因(2/3)^x>0所以只