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

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

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

\(\Rightarrow x=3\)

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

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

\(\Leftrightarrow x=3\)

có thiếu dấu ko ạ ?

có bị thiếu dấu không vậy bạn

(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

\(5^x-2-3^2=24\)

\(5^x-2-9=24\)

\(5^x-2=24+9\)

\(5^x-2=33\)

\(5^x=33+2\)

\(5^x=35\)

mà 35 = 5.7 => không có số x nào thỏa mãn điều kiện

Vậy \(x\in\left\{\varnothing\right\}\)

Mình không chắc chắn là đúng hay sai, nếu sai bạn thông cảm!!!

*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)

\(M=x^2+11xy-y^2\)

* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

Ta có : \(\frac{6}{11}x=\frac{9}{2}y\)=> \(\frac{12x}{22}=\frac{99y}{22}\)=> 12x = 99y => 4x = 33y => \(\frac{x}{33}=\frac{y}{4}\)

\(\frac{9}{2}y=\frac{15}{5}z\)=> \(\frac{45y}{10}=\frac{30z}{10}\)=> 45y = 30z => 3y = 2z => \(\frac{y}{2}=\frac{z}{3}\)

=> \(\frac{x}{33}=\frac{y}{4};\frac{y}{2}=\frac{z}{3}\)

=> \(\frac{x}{66}=\frac{y}{4};\frac{y}{4}=\frac{z}{12}\)

=> \(\frac{x}{66}=\frac{y}{4}=\frac{z}{12}\)và y - x + z = -120

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{66}=\frac{y}{4}=\frac{z}{12}=\frac{y-x+z}{4-66+12}=\frac{-120}{-50}=\frac{12}{5}\)

=> \(\hept{\begin{cases}\frac{x}{66}=\frac{12}{5}\\\frac{y}{4}=\frac{12}{5}\\\frac{z}{12}=\frac{12}{5}\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{792}{5}\\y=\frac{48}{5}\\z=\frac{144}{5}\end{cases}}\)

=5^3+3^9

=125+19683

=19808

5 x 6 : 5 x 3 + 3 x 2 x 3 x 7 

= 30 : 15 + 126 

= 2 + 126 

= 128

5x6:5x3+3x2x3x7

=5x(6:3)+3x2x3x7

=5x2+126

=10+126

=136

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

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

\(\dfrac{1}{5}\)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                                     =\(\dfrac{-29}{15}\)