a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+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

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                           \(x=-3\)

\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)

=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)

=> \(2^{x-1}.6=\frac{7}{32}\)

=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)

ĐỀ SAI RỒI BẠN !

2^x-1+5.2^x-2=2.2^x-2+5.2^x-2=2^x-2.(5+2)=2^x-2.7=7/32

=>2^x-2=7/32:7=1/32

=>x-2=-5

=>x=-3

=> x =3

a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+1=1/16=1/2⁴

=> x+1=-4=>x=-5

a. 2^x-1+ 5.2^x-1:2=7/32

=> 2^x+1.(1+5/2)=7/32

=>2^x+1.7/2=7/32

=> 2^x+1=1/16=1/2⁴

=> x+1=-4=>x=-3

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)\(4^x+4^{x+3}=4160\)

\(4^x+4^x\cdot4^3=4160\)

\(4^x\left(1+4^3\right)=4160\)

\(4^x\cdot65=4160\)

\(4^x=4160:65=64\)

\(4^x=4^3\)

\(\Rightarrow x=3\)

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 \(x\in Z\)

\(\Rightarrow x\in\left\{-102;-101;-100\right\}\)

bài a nhầm 2 dòng cuối

=>x-1=-4

=>x=-3

Bài a bạn ST làm sai

\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}+2^{x-1}.\frac{5}{2}=\frac{7}{32}\Rightarrow2^{x-1}.\left(1+\frac{5}{2}\right)=\frac{7}{32}\Rightarrow2^{x-1}.\frac{7}{2}=\frac{7}{32}\)

\(\Rightarrow2^{x-1}=\frac{7}{32}:\frac{7}{2}=\frac{7}{32}.\frac{2}{7}=\frac{1}{16}\)

\(\Rightarrow2^{x-1}=2^{-4}\Rightarrow x-1=-4\Rightarrow x=-4+1=-3\)