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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 \(x=-3\)
a, \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
=>\(2^{x-1}+\frac{5}{2}.2^{x-1}=\frac{7}{32}\)
=>\(2^{x-1}\left(1+\frac{5}{2}\right)=\frac{7}{32}\)
=>\(2^{x-1}\cdot\frac{7}{2}=\frac{7}{32}\)
=>\(2^{x-1}=\frac{1}{16}=\frac{1}{2^4}=2^{-4}\)
=>x-1=-4
=>x=-5
b, |x - 4| + |x - 10| + |x + 101| + |x + 990| + |x + 1000| = |4-x|+|10-x|+|x+101|+|x+990|+|x+1000|
Ta có: \(\left|4-x\right|\ge4-x;\left|10-x\right|\ge10-x;\left|x+990\right|\ge x+990;\left|x+1000\right|\ge x+1000\)
\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+990\right|+\left|x+1000\right|\ge4-x+10-x+x+990+x+1000\)
\(\Rightarrow\left|4-x\right|+\left|10-x\right|+\left|x+101\right|+\left|x+990\right|+\left|x+1000\right|\ge2004+\left|x+101\right|\)
\(\Rightarrow2005\ge2004+\left|x+101\right|\)
\(\Rightarrow\left|x+1\right|\le1\)
\(\Rightarrow-1\le x+101\le1\)
\(\Rightarrow-102\le x\le-100\)
Vì \(x\in Z\)
\(\Rightarrow x\in\left\{-102;-101;-100\right\}\)
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