a) 2x - 31 + 42 = - 3x - 45

2x + 11 = - 3x - 45

2x + 3x = - 45 - 11

5x = - 56

x = - 56/5

b) - 25 + 2x = 15 + 17 - x

2x + x = 32 + 25

3x = 57

x = 19

a/ x - ( 25 - x) = x - 15

=> x - 25 + x = x - 15

=> 2x - x = -15 + 25

=> x = 10

b/ x - 8 = -13 - 8

=> x = -13 - 8 + 8 = -13

c/ x - (-15) = 5 - (-19)

=> x + 15 = 5 + 19 => x = 5 + 19 - 15 = 9

d/ (x + 35) = 2x + 15 + (13 - 4)

=> x + 35 = 2x + 15 + 13 - 4

=> x - 2x = 15 + 13 - 4 - 35

=> -x = -11 => x = 11

e/ 5 - 2x = 15-(-9) - 3x

=> 5 - 2x = 15 + 9 - 3x

=> -2x + 3x = 15 + 9 - 5

=> x = 19

g/ x + 135 = 3x - 15

=> x - 3x = -15 - 135

=> -2x = -150

=> x = -150/ - 2 = 75

a) \(2x-13=7\)

\(\Leftrightarrow2x=7+13=20\)

\(\Leftrightarrow x=\frac{20}{2}=10\)

b) \(23-x=31-\left(-9\right)\)

\(\Leftrightarrow23-x=40\)

\(\Leftrightarrow x=23-40=-17\)

c) \(\left|x+13\right|-4=16\)

\(\Leftrightarrow\left|x+13\right|=16+4=20\)

\(\Leftrightarrow\left[\begin{matrix}x+13=20\Rightarrow x=7\\x+13=-20\Rightarrow x=-33\end{matrix}\right.\)

d) Vì 45 \(⋮x\) mà x > 15

\(\Rightarrow\) x = 45

a, 2x = 7 + 13 =20 => x = 20 : 2 =10

b, 23 - x = 31 + 9 = 40 => x = 23 - 40 = -17

c, x = 7 ; x = -33

d, x = 22,5

a, \(2x-13=7\)

\(2x=7+13\)

\(2x=20\)

\(x=\frac{20}{2}\)

\(x=10\)

b, \(23-x=31-\left(-9\right)\)

\(23-x=40\)

\(x=23-40\)

\(x=-17\)

c, \(\left|x+13\right|-4=16\)

\(\left|x+13\right|=16+4\)

\(\left|x+13\right|=20\)

\(\Rightarrow\left[\begin{matrix}x+13=20\\x+13=-20\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=7\\x=-33\end{matrix}\right.\)

d, \(45⋮x;x>15\)

Vì \(45⋮x\) nên x là Ư(45)

Mà \(Ư\left(45\right)=\left\{1;3;5;9;15;45\right\}\)

Vì x > 15 nên x = 45.

a, x = -8 - 3 = -11

b, 35 + x = 27 + 12 = 39

=> x = 39 - 35 = 4

c, 2^x = 31 - 15 = 16 

=> x = 4

\(a\)) \(3+x=-8\)

\(x=-8-3\)

\(x=-11\)

a.\(3+x=-8\)
\(x=-8-3\)
\(x=-11\)
b.\(\left(35+x\right)-12=27\)
\(35+x=27+12\)
\(35+x=39\)
\(x=39-35\)
\(x=4\)
c.\(2^x+15=31\)
\(2^x=31-15\)
\(2^x=16\)
\(2^x=2^4\)
\(x=4\)

câu a : 

3 + x = -8

x = -8 - 3

x = -11

câu b : 

(35 + x) - 12 = 27

35 + x = 39

x = 39 - 35

x = 4

câu c : 

2^x + 15 = 31

2^x = 31 - 15

2^x = 16

x = \(\sqrt{16}\) = 4

Giải chi tiết cho mình nha

\(\dfrac{2}{5}+\dfrac{11}{15}=\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

\(\dfrac{7}{8}-\dfrac{7}{9}=\dfrac{63}{72}-\dfrac{56}{72}=\dfrac{7}{72}\)

\(\dfrac{11}{13}\times\dfrac{26}{31}=\dfrac{22}{31}\)

\(\dfrac{16}{24}:4=\dfrac{16}{24}\times\dfrac{1}{4}=\dfrac{4}{24}=\dfrac{1}{6}\)

\(30:\dfrac{6}{5}=30\times\dfrac{5}{6}=25\)

\(\dfrac{9}{16}:4=\dfrac{9}{16}\times\dfrac{1}{4}=\dfrac{9}{64}\)

\(\dfrac{1}{2}:\dfrac{1}{3}\times\dfrac{8}{15}=\dfrac{1}{2}\times3\times\dfrac{8}{15}=\dfrac{4}{5}\)

2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5

a)(x+17):21+3=7

(x+17):21=10

x+17=210

x=93

a, ( x + 17 ) : 21 + 3  = 7
    ( x + 17 ) : 21 = 7 - 3
    ( x + 17 ) : 21 = 4
      x + 17 = 4 x 21
      x + 17 = 84
      x = 84 - 17
      x = 67
b, 3^x+3 - 13 = 230
=> 3^{x + 3}  = 230 + 13
=> 3^x+3 = 243
=> 3^x+3 = 3⁵
=> x + 3 = 5
=> x = 5 -3
=> x = 2
c, 2x - 15 = 23
    2x = 23 + 15 = 38
    x = 38 : 2
    x = 19
d, ( x + 16 ) : 31 - 3 = 7
    ( x + 16 ) : 31 = 10
      x + 16 = 10 x 31 = 310
      x = 310 - 16
      x = 294

(x+17) : 21+3 =7

(x+17) +3 =7.21

(x+17) +3= 147

(x+17) = 147-3

(x+17) = 144

x= 144-17

x= 124

1.

b) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x.1+3^x.3^2=2430\)

\(\Rightarrow3^x.\left(1+3^2\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=2430:10\)

\(\Rightarrow3^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)

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