Tìm x:\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
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K
8 tháng 3 2019
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
YN
8 tháng 3 2019
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}+5.2^{x-1}.2^3=\frac{7}{32}\)
\(\Rightarrow2^{x-1}.\left(1+5.2^3\right)=\frac{7}{32}\)
\(\Rightarrow2^{x-1}.41=\frac{7}{32}\)
\(\Rightarrow2^{x-1}=\frac{7}{1312}\)
\(\Rightarrow\) Ko có x thỏa mãn
PT
1
14 tháng 11 2016
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
19 tháng 6 2016
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
NL
2
20 tháng 4 2017
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
LT
4
21 tháng 10 2016
b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)
\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)
\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)
\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)
\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)
\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)
\(2^x=\frac{2^0}{2^3}=2^{-3}\)
\(\Rightarrow x=-3\)
21 tháng 10 2016
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 \(x=-3\)
S
28 tháng 6 2018
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 \(x\in Z\)
\(\Rightarrow x\in\left\{-102;-101;-100\right\}\)
B
17 tháng 4 2018
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}+2^{x-1}.\frac{5}{2}=\frac{7}{32}\Rightarrow2^{x-1}.\left(1+\frac{5}{2}\right)=\frac{7}{32}\Rightarrow2^{x-1}.\frac{7}{2}=\frac{7}{32}\)
\(\Rightarrow2^{x-1}=\frac{7}{32}:\frac{7}{2}=\frac{7}{32}.\frac{2}{7}=\frac{1}{16}\)
\(\Rightarrow2^{x-1}=2^{-4}\Rightarrow x-1=-4\Rightarrow x=-4+1=-3\)
ai ko bt = -3 giải rõ ràng đi
\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(2.2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(2^{x-2}.\left(5+2\right)=\frac{7}{32}\)
\(2^{x-2}.7=\frac{7}{32}\)
\(2^{x-2}=\frac{7}{32}:7\)
\(2^{x-2}=\frac{1}{32}\)
\(\Rightarrow x-2=-5\)
\(x=-5+2\)
Vậy \(x=-3\)