a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+1=1/16=1/2⁴

=> x+1=-4=>x=-5

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}+5.2^{x-1}.2^3=\frac{7}{32}\)

\(\Rightarrow2^{x-1}.\left(1+5.2^3\right)=\frac{7}{32}\)

\(\Rightarrow2^{x-1}.41=\frac{7}{32}\)

\(\Rightarrow2^{x-1}=\frac{7}{1312}\)

\(\Rightarrow\)   Ko có x thỏa mãn

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}+2^{x-1}.\frac{5}{2}=\frac{7}{32}\Rightarrow2^{x-1}.\left(1+\frac{5}{2}\right)=\frac{7}{32}\Rightarrow2^{x-1}.\frac{7}{2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}=\frac{7}{32}:\frac{7}{2}=\frac{7}{32}.\frac{2}{7}=\frac{1}{16}\)

\(\Rightarrow2^{x-1}=2^{-4}\Rightarrow x-1=-4\Rightarrow x=-4+1=-3\)

ai ko bt = -3 giải rõ ràng đi              

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(2.2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

      \(2^{x-2}.\left(5+2\right)=\frac{7}{32}\)

                   \(2^{x-2}.7=\frac{7}{32}\)

                       \(2^{x-2}=\frac{7}{32}:7\)

                       \(2^{x-2}=\frac{1}{32}\)

               \(\Rightarrow x-2=-5\)

                              \(x=-5+2\)

Vậy                           \(x=-3\)

\(2^{x-1}+5\cdot2^{x-2=\frac{7}{32}}\)

\(2^{x-1}+5\cdot2^{\frac{71}{32}}\)

\(2^{\frac{39}{32}}+23.2744\)

\(25.60194\)

=2^x-1 + 5.(2^x : 4) = 2^x-1 + 5[(2^x-1)/2]=2^x-1 + 2,5.2^x-1=3,5.(2^x-1)=7/32

từ đó suy ra dpcm

2^x-1+5.2^x-2=2.2^x-2+5.2^x-2=2^x-2.(5+2)=2^x-2.7=7/32

=>2^x-2=7/32:7=1/32

=>x-2=-5

=>x=-3

=> x =3

2^x-1+5.2^x-2=2.2^x-2+5.2^x-2=2^x-2.(5+2)=2^x-2.7=7/32

=>2^x-2=7/32:7=1/32

=>x-2=-5

=>x=-3

ĐOán thế

Hả ! Vừa thấy 732 cơ mà ! Làm mất công rồi ....

\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)

=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)

=> \(2^{x-1}.6=\frac{7}{32}\)

=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)

ĐỀ SAI RỒI BẠN !

a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+1=1/16=1/2⁴

=> x+1=-4=>x=-5

a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+1=1/16=1/2⁴

=> x+1=-4=>x=-3