tìm X, 2(x-1)+5.2(x-2)= 7/32
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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
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
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
\(2^{\left(x-1\right)}+5.2^{\left(x-2\right)}=\frac{7}{32}\)
\(\Leftrightarrow2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\)
\(\Leftrightarrow\left(2+5\right).2^{x-2}=\frac{7}{32}\)
\(\Leftrightarrow7.2^{x-2}=\frac{7}{32}\)
\(\Leftrightarrow2^{x-2}=\frac{7}{32}:7\)
\(\Leftrightarrow2^{x-2}=\frac{1}{32}=2^{-5}\)
\(\Leftrightarrow x-2=-5\)
\(\Leftrightarrow x=\left(-5\right)+2\)
\(\Leftrightarrow x=-3\)
Vậy \(x=-3\).
Chúc bạn học tốt!
\(\Leftrightarrow2.2^{x-2}+5.2^{x-2}=7.2^{-5}\Leftrightarrow7.2^{x-2}=7.2^{-5}\)
\(\Leftrightarrow x^{x-2}=2^{-5}\Leftrightarrow x-2=-5\Leftrightarrow x=-3\)
<=> \(2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\) <=> \(\left(2+5\right).2^{x-2}=\frac{7}{32}\)
<=> \(7.2^{x-2}=\frac{7}{32}\)<=> \(2^{x-2}=\frac{1}{32}=2^{-5}\) => x - 2 = -5 => x = -3