\(5^{x+2}-5^x=3.10^3\)

\(\Rightarrow5^x\left(5^2-1\right)=3000\)

\(\Rightarrow5^x.24=3000\Rightarrow5^x=125=5^3\)

\(\Rightarrow x=3\)

\(5^{x+2}-5^x=3\cdot10^3\)

\(\Leftrightarrow5^x\cdot24=3000\)

\(\Leftrightarrow x=3\)

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)

\(\dfrac{1}{5}\)x(X+\(\dfrac{1}{5}\))+\(\dfrac{2}{5}\)x(X+\(\dfrac{5}{3}\))           = \(\dfrac{98}{75}\)
=> \(\dfrac{1}{5}\)X+\(\dfrac{2}{5}\)+\(\dfrac{6}{15}\)X+\(\dfrac{31}{15}\)           = \(\dfrac{98}{75}\)
=> (\(\dfrac{1}{5}\)X+\(\dfrac{6}{15}\)X)+(\(\dfrac{2}{5}\)+\(\dfrac{31}{15}\))      =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{1}{5}\)+\(\dfrac{6}{15}\))+(\(\dfrac{6}{15}\)+\(\dfrac{31}{15}\))    =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{3}{15}\)+\(\dfrac{6}{15}\))+\(\dfrac{37}{15}\)            = \(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)+\(\dfrac{37}{15}\)                      =\(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)                             =\(\dfrac{98}{75}\)-\(\dfrac{37}{15}\)
=>X                                     =\(\dfrac{-29}{25}\):\(\dfrac{9}{15}\)
=>X                                     =\(\dfrac{-29}{15}\)

 \(\dfrac{5}{7}< \dfrac{x}{5}< 1\) 

\(\dfrac{5\times5}{7\times5}< \dfrac{7\times x}{7\times5}< \dfrac{35}{35}\)

\(\dfrac{25}{35}< \dfrac{7\times x}{35}< \dfrac{35}{35}\)

\(25< 7\times x< 35\)

\(\dfrac{25}{7}< x< \dfrac{35}{7}\)

\(\dfrac{25}{7}< x< 5\)

\(3< x< 5\) (x is the natural number)  

x=4 

 answer please

a: \(=\dfrac{7}{5}\cdot\dfrac{15}{49}-\dfrac{12+10}{15}:\dfrac{11}{5}\)

\(=\dfrac{3}{7}-\dfrac{22}{15}\cdot\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=\dfrac{-5}{21}\)

b: =>2,8x-32=-60

=>2,8x=-28

hay x=-10

/x+5/+/x-1/=12:/y+5/+3

suy ra x+5+x-1=12:/y+5/+3 hoặc x+5+x-1=12:-<y+5>+3

tính cái gì bạn

\(2^x+2^{x-1}+2^{x-2}=2^{x-2}.2^2+2^{x-2}.2+2^{x-2}\)\(=2^{x-2}\left(4+2+1\right)=2^{x-2}.7\)

thank you

(x – 2) – (39 – 45) = | - 5| - | -10|

(x - 2) - 6 = 5 - 10

(x - 2) - 6 = - 5

 x - 2 = - 5  + 6

 x - 2 = 1

       x = 1 + 2

       x = 3

Vậy x = 3

\(\left(x-2\right)+6=-5\)

\(\Rightarrow x-2=-11\)

\(\Rightarrow x=-9\)

( x - 2 ) - ( 39 - 45 ) = | -5 | - | -10 |

( x -2 ) -    (- 6) = 5 - 10

( x - 2 ) + 6    = -5

 x - 2      = -5 - 6

x - 2        =  -11

x             = -11 + 2

x              = -9

Mình viết nhầm đề, xin lỗi mọi người :

|-x -5| = |1-5|

Ta có: \(|-x-5|=|1-5|\)

<=> \(|-x-5|=4\)

<=>\(\orbr{\begin{cases}x-5=4\\x-5=-4\end{cases}}\)

<=>\(\orbr{\begin{cases}x=9\\x=1\end{cases}}\)

Vậy x=1 hoặc x=9

=>-x-5=4 hoặc -x-5=-4

=>x=-9 hoặc x=1

chúc bạn học tốt!nhớ