( x - 15 ) x 7 - 270 : 45 = 169

          ( x - 15 ) x 7 - 6  = 169

          ( x - 15 ) x 7       = 175

            x - 15               = 25

                  x                = 40

( x-15 ) x7 - 270 : 45 = 169

( x - 15) x 7               =169  x 45

( x - 15) x 7               =7605

x - 15                       =7605 : 7

x - 15                      =7620

        x                       =7620 + 15

       x                        

x=40 nha bạn !!!                                                                                                                                                                                                   chúc bạn hok tốt nha.................

a) -32 -4.(x-5) = 0

<=>4.(x-5)=-32

<=>x-5=(-32):4

<=>x+5=-8

<=>x=-8+5

<=>x=-3

Vậy x=-3

b) 13.(x-5)=-169

<=>x-5=(-169):13

<=>x-5=-13

<=>x=-13+5

<=>x=-8

vậy x=-8

c) (-2).x+5 = (-3).(-3)+8

<=>(-2).x+5=17

<=>(-2).x=17-5

<=>(-2).x=12

<=>x=12:(-2)

<=>x=-6

Vậy x=-6

d) (-8).x = (-10).(-2)-4

<=>(-8).x=16

<=>x=16:(-8)

<=>x=-2

vậy x=-2

e) (-9).x+3 = (-2).(-7)+16

<=>(-9).x+3=30

<=>(-9).x=30-3

<=>(-9).x=27

<=>x=27:(-9)

<=>x=-3

Vậy x=-3

a) x - 15 = 7 + (-2)

x-15 = 5

x=20

b) | x - 5 | = 15 + (-8)

I x - 5 I = 7

1) x-5 = 7 => x = 12

2) x-5 = -7 => x= -2

Vậy x=12 hoặc x=-2

c) I x | -  25 = 45 + (-15) 

 I x | -  25 = 30

I xI = 55

Vậy x= 55 hoặc x=-55

a) x - 15 = 7 + ( - 2 )

x - 15 = 5

      x  = 5 + 15

      x  = 20

b) | x - 5 | = 15 + ( - 8 )

| x - 5 | = 7

=> x = 7 + 5 

     x = 12

c) | x | - 25 = 45 + ( - 15 )

| x | - 25 = 30

=> | x | = 30 + 25

     | x | = 55

=> x = 55

phần cuối x = - 55 hoặc 55

Mình viết thiếu

Xin lỗi nha

a) (x - 15) × 7 - 270 : 45 = 169

(x - 15) × 7 - 6 = 169

(x - 15) × 7 = 169 + 6

(x - 15) × 7 = 175

x - 15 = 175 : 7

x - 15 = 25

x = 25 + 15

x = 40

b) [(4x + 28) × 3 + 55] : 5 = 35

(4x + 28) × 3 + 55 = 35 × 5

(4x + 28) × 3 + 55 = 175

(4x + 28) × 3 = 175 - 55

(4x + 28) × 3 = 120

4x + 28 = 120 : 3

4x + 28 = 40

4x = 40 - 28

4x = 12

x = 12 : 4

x = 3

c) (455 × x : 2 × 6) : 5 = 31

455 × x : 2 × 6 = 31 × 5

455 × x : 2 × 6 = 155

x × 455 : 2 × 6 = 155

x × 1365 = 155

x = 155 : 1365

x = 31/273

d) 128 × x - 12 × x - 16 × x = 520800

(128 - 12 - 16) × x = 520800

100 × x = 520800

x = 520800 : 100

x = 5208

e) (x × 0,25 + 2022) × 2023 = (50 + 2022) × 2023

(x × 0,25 + 2022) × 2023 = 2072 × 2023

(x × 0,25 + 2022) × 2023 = 4191656

x × 0,25 + 2022 = 4191656 : 2023

x × 0,25 + 2022 = 2072

x × 0,25 = 2072 - 2022

x × 0,25 = 50

x = 50 : 0,25

x = 200

f) 4 × x + 100 = x + 280

4 × x - x = 280 - 100

(4 - 1) × x = 180

3 × x = 180

x = 180 : 3

x = 60

g) (x + 1) + (x + 2) + (x + 3) + ... + (x + 100) = 7450

x + 1 + x + 2 + x + 3 + ... + x + 100 = 7450

100 × x + 100 × 101 : 2 = 7450

100 × x + 5050 = 7450

100 × x = 7450 - 5050

100 × x = 2400

x = 2400 : 100

x = 24

a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)

\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)

=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)

b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)

Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)

c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)

\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)

=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)

d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)

\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6

=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)

a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)

b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)

c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)

\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)

d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)

\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)

Mong bạn xem lại đề bài.

\(a,78-\left(x-3\right)=16\)

\(78-x+3=16\)

\(51-x=16\)

\(x=51-16=35\)

\(b,\left(2x+3\right):4=60\)

\(2x+3=60\cdot4=240\)

\(2x=240-3=237\)

\(x=\frac{237}{2}\)

\(c,\left(2x-8\right)=\left(2x-8\right)^7\)

\(\left(2x-8\right)-\left(2x-8\right)^7=0\)

Để mik nghĩ thêm tí r làm tiép

Để tui làm nốt ý C cho !!!

( 2x - 8 ) = ( 2x - 8 )⁷

=> ( 2x - 8 ) . 1 - ( 2x - 8 ) . ( 2x - 8 )⁶ = 0

=> ( 2x - 8 ) . [ 1 - ( 2x - 8 )⁶ ] = 0

=> 1 - ( 2x - 8 )⁶ = 0

=> ( 2x - 8 )⁶ = 1

=> ( 2x - 8 )⁶ = ( 1⁶ )

\(\Rightarrow\orbr{\begin{cases}2x-8=1\\2x-8=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=9\\2x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{7}{2}\end{cases}}}\)

Vậy ........

c) \(\left(2x-8\right)=\left(2x-8\right)^7\)

\(\Leftrightarrow\left(2x-8\right)^7-\left(2x-8\right)=0\)

\(\Leftrightarrow\left(2x-8\right)\left[\left(2x-8\right)^6-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-8=0\\\left(2x-8\right)^6-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(2x-8\right)^6=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x\in\left\{\frac{7}{2};\frac{9}{2}\right\}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{7}{2};\frac{9}{2}\right\}\)

\(a,\Leftrightarrow x-28=-45\\ \Leftrightarrow x=-27\\ b,\Leftrightarrow3+x=0\\ \Leftrightarrow x=-3\\ c,\Leftrightarrow\left[{}\begin{matrix}7-x=0\\-x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ d,\Leftrightarrow16\left(x^2-4\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

a)-28+x=-34+(-11)             b)-12(3+x)=0

<=>-28+x=-45                   <=>-36-12x=0

<=>x=-17                            <=>-12x=36

Vậy x=-17                             <=>x=-3

                                         Vậy x=-3

c)(7-x)(-x+2)=0

<=>7-x=0 hoặc -x+2=0

Th1:7-x=0                    Th2:-x+2=0

 <=>x=7                             <=>x=2

           Vậy xϵ{7;2}

d)16x2-64=0

<=>32x=64

<=>x=2

Vậy x=2