- Mình có thể giúp bài 1 :)

 5^x +  5^ ( x + 2 ) = 650

5^x +  5^x . 5² = 650

 5^x .( 1 + 25 ) = 650

 5^x . 26 = 650 

5^x = 650 : 26

 5^x = 25

 5^x = 5² 

=> x = 2 

Vậy x = 2 

The Reflection Of Light~ Ủa đề  là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy

7x-2x=6^17:6^15+44:11

=>2x=6^(17-15)+4

=>2x=6^2+4

=>2x=36+4

=>2x=40

=>x=40:2

=>x=20

5x=6^2+4

5x=40

x=40:5=8

Vậy x=8

7x - 2x = 6^17 : 6^15 + 44 : 11

5x         = 6^2 + 4

5x          = 36+4

5x            = 40

x              = 8

bạn có thể check lại đề bài câu a được không ạ

là x^2 = x^5 á bạn

a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\) 

    \(x\)        = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)

    \(x\)        = \(\dfrac{52}{63}\)

b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)

     2\(x\).(2\(x\) + 1) = 30

     4\(x^2\)+ 2\(x\)  - 30 = 0

  4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0

   (4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0

   4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0

        2 (\(x\) + 3).(2\(x\) -  5) = 0

            \(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)

             \(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}

c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\) = \(\dfrac{85}{x}\)

     \(\dfrac{85}{x}\)         = \(\dfrac{85}{24}\)

       \(x\)           = 24

\(7x-2x=6^{17}:6^{15}+44:11\)

\(\Rightarrow5x=6^2+4\)

\(\Rightarrow5x=36+4=40\)

\(\Rightarrow x=40:5\)

\(\Rightarrow x=8\)

7x-2x=6¹⁷:6¹⁵+44:11

5x=36+4

5x=40

x=40:5

x=8

vậy: x=8

a)\(\left(\frac{9}{2}-2x\right)\left(\frac{-11}{2}\right)=\frac{-11}{15}\)

    \(-\frac{99}{4}-\left(-11x\right)=\frac{-11}{15}\)

    \(-\frac{99}{4}+11x=\frac{-11}{15}\)

    \(11x=\frac{-11}{15}-\left(-\frac{90}{4}\right)\)

     \(11x=\frac{653}{30}\)

      \(x=\frac{653}{30}:11\)

     \(x=\frac{653}{330}\)

Vậy \(x=\frac{653}{330}\)

b)\(\left(x-\frac{5}{6}\right):\frac{9}{21}=-\frac{7}{5}\)

    \(\left(x-\frac{5}{6}\right)=-\frac{7}{5}.\frac{9}{21}\)

      \(x-\frac{5}{6}=-\frac{3}{5}\)

     \(x=\frac{7}{30}\)

Vậy \(x=\frac{7}{30}\)

c)2x+3x=-15

   5x=-15

   x=-15:5

   x=-3

Vậy x=-3

a) \(6x+15\times8=12\times\left(19-x\right)\)

\(6x+120=228-12x\)

\(6x+120-228+12x=0\)

\(18x-108=0\)

\(18x=108\)

\(x=6\)

b) \(160-\left(35\div x+3\right)\times15=15\)

\(160-\left(35\div x+3\right)=1\)

\(35\div x+3=159\)

\(35\div x=156\)

\(x=\dfrac{35}{156}\)

c) \(2x-\left(1309\div11-19\right)-2=0\)

\(2x-1309\div11-19=2\)

\(2x-119-19=2\)

\(2x-119=21\)

\(2x=140\)

\(x=70\)

d) \(\left(x-7\right)\times\left(2x-16\right)=0\)

\(x-7=0;2x-16=0\)

\(x=7;2x=16\)

\(x=7;x=8\)

a) $6x+15\times 8=12\times \left(19-x\right)$

$6x+120=228-12x$

$6x+120-228+12x=0$

$18x-108=0$

$18x=108$

$x=6$

b) $160-\left(35÷x+3\right)\times 15=15$

$160-\left(35÷x+3\right)=1$

$35÷x+3=159$

$35÷x=156$

�=35156x=15635​