(\(x-3\))²⁰ = (\(x-3\))¹⁰

(\(x-3\))²⁰  - (\(x-3\))¹⁰ = 0

(\(x-3\))¹⁰.[( \(x-3\))¹⁰ - 1] = 0

\(\left[{}\begin{matrix}\left(x-3\right)^{10}=0\\\left(x-3\right)^{10}-1=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x-3=\pm1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

Vậy \(x\in\) { 2; 3; 4}

    n + 2 ⋮ n - 4

⇔ n - 4 + 6 ⋮ n - 4

⇔ 6 ⋮ n - 4

n - 4 \(\in\) Ư( -6; -3; -2; -1; 1; 2; 3; 6}

n \(\in\) { -2; 1; 2; 3; 5; 6; 7; 10}

cần gấp 

\(=2017.2018.2023.\left(58-58\right)=0\)

a, 7x + 10x  = 5x 

    17x = 5x

17x - 5x = 0

      12x = 0

          x =0

2; 

a, 4x + 7x = 22

    11x = 22

        x = 2

b, 12x - 8x = 25

     4x = 25

       x = \(\dfrac{25}{4}\)

c,  \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)x = \(\dfrac{4}{5}\) 

     (\(\dfrac{1}{2}-\dfrac{1}{3}\))x = \(\dfrac{4}{5}\)

    \(\dfrac{1}{6}\)x     = \(\dfrac{4}{5}\) 

      x = \(\dfrac{4}{5}\) : \(\dfrac{1}{6}\)

     x = \(\dfrac{24}{5}\)

  100 + 98 + 96 +....+ 4 + 2 - 97 - 95 -...- 3- 1

= 100 + ( 98 - 97 ) + ( 96 - 95) +...+ (2- 1)

= 100 + 1 + 1 +...+ 1       ( có 49 hạng tử là 1)

= 100 + 1  x  49

= 100 + 49

= 149 

nhanh

Tính phải k nhỉ?

`1)`

`2x + 3x + 5x`

`= (2 + 3 + 5)x`

`= 10x`

`2)`

`2.x - x + 3.x`

`= (2 - 1 + 3)x`

`= 4x`

`3)`

`9.x - 3 - 3.x`

`= (9 - 3)x - 3`

`= 6x - 3`

`4)`

Thiếu dấu, bạn bổ sung thêm

`5)`

`x - 0,2x - 0,1x`

`= (1 - 0,2 - 0,1)x`

`=0,7x`

`6)`

\(\dfrac{7}{2}x-\dfrac{1}{2}x=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)x=3x\)

`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

`@` `\text {Ans}`

`\downarrow`

`1)`

\(x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow` \(x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{7}{6}\)

Vậy, `x =`\(\dfrac{7}{6}\)

`2)`

\(\dfrac{3}{5}-x=\dfrac{1}{3}\)

`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)

`\Rightarrow`\(x=\dfrac{4}{15}\)

Vậy, `x =`\(\dfrac{4}{15}\)

`3)`

\(\dfrac{3}{4}+x=\dfrac{7}{2}\)

`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)

`\Rightarrow`\(x=\dfrac{11}{4}\)

Vậy, \(x=\dfrac{11}{4}\)

`4)`

\(x-\dfrac{4}{3}=\dfrac{7}{9}\)

`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)

`\Rightarrow`\(x=\dfrac{19}{9}\)

Vậy, `x=`\(\dfrac{19}{9}\)

`5)`

\(x-\dfrac{5}{6}=\dfrac{7}{3}\)

`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)

`\Rightarrow x =`\(\dfrac{19}{6}\)

Vậy, `x=`\(\dfrac{19}{6}\)

`6)`

\(x-\dfrac{1}{5}=\dfrac{9}{10}\)

`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)

`\Rightarrow x=`\(\dfrac{11}{10}\)

Vậy, `x=`\(\dfrac{11}{10}\)

1; 5.2² + (\(x\) + 3) = 5²

   5.4  +  (\(x\) + 3) = 25

    20 + (\(x\) + 3) = 25

             \(x\) + 3 = 25 - 20

             \(x+3\) = 5

             \(x\)       = 5  - 3

            \(x\)        = 2

            Vậy \(x=2\)

2; 2³ + (\(x\) - 3²) = 5³ - 4³

   8 +  (\(x\) - 9)    = 125 - 64

   8 + (\(x\) - 9) = 61

         \(x\) - 9 = 61 - 8

         \(x\) - 9 = 53

        \(x\)        = 53  + 9

        \(x\)       = 62

        Vậy \(x\) = 62