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

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-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

a, x = 384

b, x = 161

c, x = 8

d, x = 15

X

ko biet

1/5.2^x+1/3.2^x.2=1/5.2^7+1/3.2^7.2

2x(1/5+1/3.2)=2^7(1/5+1/3.2)

=>2^x=2^7
=> x=7

a/ \(\frac{2}{3}.3^{x+1}-7.3^x=405\)

<=> 2.3^x-7.3^x=-405

<=> 5.3^x=405

<=> 3^x=81 = 3⁴

=> x=4

b/ (0,4x-1,3)²=5,29=(2,3)²

=> \(\hept{\begin{cases}0,4x-1,3=2,3\\0,4x-1,3=-2,3\end{cases}}\)=> \(\hept{\begin{cases}x=9\\x=-\frac{5}{2}\end{cases}}\)

c/ 5.2^x+1.2^-2-2^x=384

<=> 5.2^x-1-2.2^x-1=384

<=> 3.2^x-1=384

<=> 2^x-1=128=2⁷

=> x-1=7 => x=8

d/ 3^x+2.5^y=45^x

<=> 3^x+2.5^y=3^2x.5^x

=> \(\hept{\begin{cases}x+2=2x\\x=y\end{cases}}\)=> x=y=2

\(\Leftrightarrow2.2^{x-2}+5.2^{x-2}=7.2^{-5}\Leftrightarrow7.2^{x-2}=7.2^{-5}\)

\(\Leftrightarrow x^{x-2}=2^{-5}\Leftrightarrow x-2=-5\Leftrightarrow x=-3\)

<=> \(2.2^{x-2}+5.2^{x-2}=\frac{7}{32}\) <=> \(\left(2+5\right).2^{x-2}=\frac{7}{32}\)

<=> \(7.2^{x-2}=\frac{7}{32}\)<=> \(2^{x-2}=\frac{1}{32}=2^{-5}\) => x - 2 = -5 => x = -3

2^x-1+5.2^x-2=7/32

=>2.2^x-2+5.2^x-2=7/32

=7.2^x-2=7/32

=>2^x-2=1/32

=>2^x-2=2^-5

=>x-2=-5

=>x=-3

vậy x=-3