|x-3,4|>0

|x-2,6|>0

=>|x-3,4|+|x-2,6|>0

mà theo đề:|x-3,4|+|x-2,6|=0

=>|x-3,4|=|x-2,6|=0

=>x=3,4;x=2,6

ko thể tồn tại 2 giá trị x cùng lúc

=>x thuộc rỗng

tick nhé

\(\text{a) Ta co }\) \(4^{x+3}-3.4^{x+1}=13.4^{11}\)

\(\Rightarrow\)       \(4^{x+1}\left(16-3\right)=13.4^{11}\)

\(\Rightarrow4^{x+1}.13=13.4^{11}\)

\(\Rightarrow4^{x+1}=4^{11}\)

\(\Rightarrow x+1=11\)

\(\Rightarrow\text{x=10}\)

a) 

\(4^{x+3}-3.4^{x+1}=13.4^{11}\)

<=> \(4^{x+1}\left(16-3\right)=13.4^{11}\)

<=> \(4^{x+1}.13=13.4^{11}\)

<=> \(4^{x+1}=4^{11}\)

<=> \(x+1=11\)

<=> x=10

a: =>(x-2)^3*[(x-2)^2-1]=0

=>(x-2)(x-3)(x-1)=0

=>\(x\in\left\{1;2;3\right\}\)

b: =>(x-3)^2*(x-3-1)=0

=>(x-3)(x-4)=0

=>x=3 hoặc x=4

c: =>\(11\cdot\dfrac{6^x}{6}+2\cdot6^x\cdot6=6^{11}\left(11+2\cdot6^2\right)\)

=>6^x(11/6+12)=6^12(11/6+12)

=>x=12

X=0

cho minh nha

b)11.6^x-1+2.6^x+1=11.6¹¹+2.6¹³

11.6^x-1+2.6^x+1 = 11. 6^12-1+ 2. 6¹²⁺¹

=> x= 12

c) 2^4-x / 16⁵ = 32⁶

2^4-x / 2²⁰= 2³⁰

2^4-x = 2⁵⁰

=> 4-x = 50

x= -46

c)      \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

       \(2^{4-x}:2^{20}=\left(2^5\right)^6\)

       \(2^{4-x}=2^{30}.2^{20}\)

       \(2^{4-x}=2^{50}\)

=>    \(4-x=50\)

=>       \(x=4-50=-46\)

vậy x = -46