2\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+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
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}+2^{x-1}.\frac{5}{2}=\frac{7}{32}\Rightarrow2^{x-1}.\left(1+\frac{5}{2}\right)=\frac{7}{32}\Rightarrow2^{x-1}.\frac{7}{2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}=\frac{7}{32}:\frac{7}{2}=\frac{7}{32}.\frac{2}{7}=\frac{1}{16}\)
\(\Rightarrow2^{x-1}=2^{-4}\Rightarrow x-1=-4\Rightarrow x=-4+1=-3\)
\(2^{x-1}+5\cdot2^{x-2=\frac{7}{32}}\)
\(2^{x-1}+5\cdot2^{\frac{71}{32}}\)
\(2^{\frac{39}{32}}+23.2744\)
\(25.60194\)
2x-1+5.2x-2=2.2x-2+5.2x-2=2x-2.(5+2)=2x-2.7=7/32
=>2x-2=7/32:7=1/32
=>x-2=-5
=>x=-3
2x-1+5.2x-2=2.2x-2+5.2x-2=2x-2.(5+2)=2x-2.7=7/32
=>2x-2=7/32:7=1/32
=>x-2=-5
=>x=-3
ĐOán thế
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5