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Ta có: \(2^{x-1}+2^{x+1}=5.2^4\)
\(\Rightarrow2^x\left(\frac{1}{2}+2\right)=5.2^4\)
\(\Rightarrow2^x.\frac{5}{2}=5.2^4\)
\(\Rightarrow2^x=2^4.2=2^5\Rightarrow x=5.\)
a: \(\Leftrightarrow2^x\cdot2\cdot2^{-2}\cdot5=384\)
\(\Leftrightarrow2^x\cdot\dfrac{5}{2}=384\)
\(\Leftrightarrow2^x=153,6\)(vô lý)
b: Sửa đề:\(3^{x+2}\cdot5^y=45^x\)
\(\Leftrightarrow3^{x+2}\cdot5^y=3^{2x}\cdot5^x\)
=>x=y và 2x=x+2
=>x=y=2
2x-1+5.2x-2=7/32
=>2.2x-2+5.2x-2=7/32
=7.2x-2=7/32
=>2x-2=1/32
=>2x-2=2-5
=>x-2=-5
=>x=-3
vậy x=-3
k nha
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\)
\(=7.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^{x-2}=\frac{1}{32}\)
\(=2^{x-2}=2^{-5}\)
\(\Rightarrow x-2=-5\)
\(\Rightarrow x=-3\)
\(\Leftrightarrow2.2^{x-2}+5.2^{x-2}=7.2^{-5}\Leftrightarrow7.2^{x-2}=7.2^{-5}\)
\(\Leftrightarrow x^{x-2}=2^{-5}\Leftrightarrow x-2=-5\Leftrightarrow x=-3\)
<=> \(2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\) <=> \(\left(2+5\right).2^{x-2}=\frac{7}{32}\)
<=> \(7.2^{x-2}=\frac{7}{32}\)<=> \(2^{x-2}=\frac{1}{32}=2^{-5}\) => x - 2 = -5 => x = -3
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
a)\(\Rightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)
b)\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(VL\right)\\x^2=4\Rightarrow x=2,-2\end{cases}}}\)VL là vô lý do bình phương luôn là số dương
Ủng hộ minhf bằng cachs k đúng nha
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}+2^{x-1}.\frac{5}{2}=\frac{7}{32}\Rightarrow2^{x-1}.\left(1+\frac{5}{2}\right)=\frac{7}{32}\Rightarrow2^{x-1}.\frac{7}{2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}=\frac{7}{32}:\frac{7}{2}=\frac{7}{32}.\frac{2}{7}=\frac{1}{16}\)
\(\Rightarrow2^{x-1}=2^{-4}\Rightarrow x-1=-4\Rightarrow x=-4+1=-3\)
5 . 2 x - 1 - 2 x + 1 = - 192
2 x ( - 5 - 1 ) = - 192
2 x . ( - 6 ) = - 192
2 x = 32
2 x = 2 5
=> x = 5