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\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\Leftrightarrow\frac{2^x}{2}+5.\frac{2^x}{2^2}=\frac{7}{32}\Leftrightarrow2^x\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\Leftrightarrow2^x=\frac{1}{8}=2^{-3}\)
<=> x=-3
`Answer:`
`\frac{x-1}{34}+\frac{x-2}{34}=\frac{x-3}{32}+\frac{x-4}{31}`
`<=>\frac{x-1}{34}-1+\frac{x-2}{33}-1=\frac{x-3}{32}-1+\frac{x-4}{31}-1`
`<=>\frac{x-35}{34}+\frac{x-35}{33}=\frac{x-35}{32}+\frac{x-35}{31}`
`<=>(x-35)(\frac{1}{34}+\frac{1}{33}-\frac{1}{32}-\frac{1}{31})=0`
`<=>x-35=0`
`<=>x=35`
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
a) \(\left(\frac{1}{2}\right)^x=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
=> x = 5
b) \(\left(\frac{5}{7}\right)^x=\frac{125}{343}\)
\(\left(\frac{5}{7}\right)^x=\left(\frac{5}{7}\right)^3\)
=> x = 3
Từ đè bài ta có :
\(\left(x+1\right)^2\)= -2 . (-32)
\(\left(x+1\right)^2\)= 64
\(\left(x+1\right)^2\)= \(^{8^2}\)= \(\left(-8\right)^2\)
=> x + 1 = 8 hoặc x + 1 =-8
x = 7 ; x = -9