( 7.x - 15 ) : 3 = 2

7.x - 15 = 6

7.x = 6 + 15

7.x = 21

  x = 3

không biết vì chưa học

88 - 3 ( 7 + x ) = 64

3 ( 7 + x ) = 88 - 64

3 ( 7 + x ) = 24

7 + x = 8

      x = 1

(7.x -15):3=2

7.x-15=2.3

7.x-15=6

7.x=6+15

7.x=21

x=21:7

x=3

12.(x+37)=504

x+37=504:12

x+37=42

x=42-37

x=5

88-3(7+x)=64

3(7+x)=88-64

3(7+x)=24

7+x=24:3

7+x=8

x=8-7

x=1

dấu . là nhân nhé

a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50

b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1

c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10

d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13

ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0

1.Tính 

a) 7.[ 6 : 2 - 15 :(-3) - l-3l ]

=  7.[ 3 + 5 - 3]

=  7.[( 3 - 3 ) + 5]

=  7.[0 + 5]

=  7.5

=  35

b) 159.(18-59) - 59 .(18-159)

= 159 . ( - 41) - 59 . ( - 141 )

= ( - 6519 ) - ( - 8319 )

= 1800

2.Tìm x thuộc Z

a) x + 15 = 20 -4x

    x+4x=20-15

    5x=5

      x=5:5

      x=1

Vậy x=1

b) 3 - lx - 1l =0

    |x-1|=3

* x-1=3            * x-1=-3

      x=3+1             x=-3+1

      x=4                 x=-2

Vậy x=4 hoặc x=-2

c) 7(x-3) - 5 (3-x) = 11x - 5

    7x-21-15+5x=11x-5

    -21-15+5=11x-7x-5x

    -31=-x

     31=x

Vậy x=31  

a, 3.(7+x)=88-64

3.(7+x)=24

7+x=24:3

7+x=8

x=1

b,7x=114-44

7x=70

x=70:7

x=10

c,31x-41=114

31x=114+41

31x=155

x=5

k như đã nói nha 3 cái nha

     3.(7+x)=88-64                           7x=114-44.                      31x-41=57.2

     3.(7+x)=24                                7x=70                              31x-41=114

         7+x=24:3                                 x=70:7                           31x     =114+41

         7+x=8                                      x=10                              31x     =155

             x=8-7                                                                             x     =155:31

                                                                                                   x     = 5

            x=1

a) \(2^x\)\(-\)\(64\)\(=\)\(2^6\)

    \(2^x\)\(-\)\(2^6\)\(=\)\(2^6\)

           \(2^x\)          \(=\)\(2^6\)\(+\)\(2^6\)

           \(2^x\)          \(=\) \(64\)\(+\)\(64\)

            \(2^x\)         \(=\)        \(128\)

  \(\Rightarrow\) \(2^x\)          \(=\)         \(2^7\)

b) \(\left(7x-11\right)\)\(^3\)\(=\)\(2^5\)\(.\)\(5^2\)\(+\)\(200\)

    \(\left(7x-11\right)\)\(^3\)\(=\)\(32\)\(.\)\(25\)\(+\)\(200\)

    \(\left(7x-11\right)\)\(^3\)\(=\)           \(1000\)

    \(\left(7x-11\right)\)\(^3\)\(=\)            \(10^3\)

\(\Rightarrow\)\(7x-11\)\(=\)\(10\)

               \(7x\)       \(=\)\(10+11\)

               \(7x\)        \(=\)    \(21\)

                  \(x\)        \(=\) \(21\)\(:\)\(7\)

                 \(x\)         \(=\)      \(3\)

\(2^x-64=2^6\)

\(2^x-64=64\)

\(2^x=64+64\)

\(2^x=128\)

\(\)Vì \(128=2^7\) \(\Rightarrow x=7\)

a, 2^x-64=2⁶

=> 2^x-64=64

=> 2^x=64-64

=> 2^x=0

=> x=0

b, (7x-11)³=2⁵.5²+200

=> (7x-11)³=32.25+200

=> (7x-11)³=1000

=> (7x-11)³=10³

=> 7x-11=10

=> 7x=10+11

=> 7x=21

=> x=21:7

=> x=3

**** ko z, t đag cần 

a) \(2^x=16\)

\(2^x=2^4\)

\(\Rightarrow x=4\)

vay \(x=4\)

b) \(4^x=64\)

\(4^x=4^3\)

\(\Rightarrow x=3\)

vay \(x=3\)

c) \(15^x=225\)

\(15^x=15^2\)

\(\Rightarrow x=2\)

vay \(x=2\)

d) \(2^x-15=17\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

vay \(x=5\)

e) \(2^x-64=2^6\)

\(2^x=2^6+64\)

\(2^x=128\)

\(2^x=2^7\)

\(\Rightarrow x=7\)

vay \(x=7\)

f) \(\left(7x-11\right)^3=2^5.5^2+200\)

\(\left(7x-11\right)^3=800+200\)

\(\left(7x-11\right)^3=1000\)

\(\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

vay \(x=3\)

a, x=4

b, x=3

c, x=2

d, =2

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2