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\(\Leftrightarrow2.2^{x-2}+5.2^{x-2}=7.2^{-5}\Leftrightarrow7.2^{x-2}=7.2^{-5}\)
\(\Leftrightarrow x^{x-2}=2^{-5}\Leftrightarrow x-2=-5\Leftrightarrow x=-3\)
<=> \(2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\) <=> \(\left(2+5\right).2^{x-2}=\frac{7}{32}\)
<=> \(7.2^{x-2}=\frac{7}{32}\)<=> \(2^{x-2}=\frac{1}{32}=2^{-5}\) => x - 2 = -5 => x = -3
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
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\)
\(=7.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^{x-2}=\frac{1}{32}\)
\(=2^{x-2}=2^{-5}\)
\(\Rightarrow x-2=-5\)
\(\Rightarrow x=-3\)
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
b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)
\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)
\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)
\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)
\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)
\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)
\(2^x=\frac{2^0}{2^3}=2^{-3}\)
\(\Rightarrow x=-3\)
a) \(4^x+4^{x+3}=4160\)
\(\Rightarrow4^x+4^x.4^3=4160\)
\(\Rightarrow4^x.\left(1+4^3\right)=4160\)
\(\Rightarrow4^x.65=4160\)
\(\Rightarrow4^x=64\)
\(\Rightarrow4^x=4^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)
\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)
\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)
\(\Rightarrow2^x=\frac{1}{8}\)
\(\Rightarrow2^x=2^{-3}\)
\(\Rightarrow x=-3\)
Vậy \(x=-3\)
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
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
Nếu như đề đúng như trên không sai xót thì mình giải như sau
2x-3+ 5.2x-2 = 7/32
2x-2 : 2 + 5.2x-2 = 7/32
2x-2 . 1/2 + 5.2x-2 = 7/32
2x-2.(5+1/2) = 7/32
2x-2. 11/2=7/32
2x-2 = 7/32 : 11/2
2x-2 = 7/176
=> Không có giá trị x thỏa mãn
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