a ) 10 x X - 1 - 3 - 5 - 7 - ... - 19 = 2 + 4 + 6 + ... + 20

10 x X - 1 - 3 - 5 - 7 - ... - 19 = 110

10 x X - ( 1 + 3 + 5 + 7 + ... + 19 ) = 110

10 x X - 100 = 110

10 x X = 110 + 100

10 x X = 210

       X = 210 : 10

       X = 21

a 10 x X-1-3-5-7-....-19 = 2+4+6+....+20

​10xX-1-3-5-7-....-19=110

​10xX=110+1+3+5+7+....+19

​10xX=210

​X=210:10

​X=21

b là 4

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a)   | x + 10 | - ( 5 - 3x ) = ( 4x - 10 ) - ( x - 5 )

=>                   | x + 10 | = ( 5 - 3x ) + ( 4x - 10 ) - ( x - 5 )

=>                   | x + 10 | = 5 - 3x + 4x - 10 - x + 5

=>                   | x +10 | = 0

=>                     x + 10 = 0

=>                      x = -10

              Vậy...

b) Làm tương tự 

Kết quả : | 3x + 21 | = -10  ( vô lí)       ( vì |3x+21| >= 0 mà -10<0)

Vậy không tìm được x thỏa mãn bài toán

\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\\ \Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^{10}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)

\(\Rightarrow x=0\)

Thử \(\left(3x-1\right)^{10}=\left(3.0-1\right)^{10}=\left(-1\right)^{10}=1\)

\(\left(3x-1\right)^{20}=\left(3.0-1\right)^{20}=\left(-1\right)^{20}=1\)

Suy ra \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

Đề bài yêu cầu tìm gì thế?

\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

\(\Rightarrow\left(3x-1\right)^{10}=\left(3x-1\right)^{20}=0\)

\(\Rightarrow\left(3x-1\right)^{10}.\left[\left(3x-1\right)^{10}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-1=0\\3x-1=\pm0\end{cases}}}\)

\(\left(+\right)3x-1=0\Rightarrow x=\frac{1}{3}\)

\(\left(+\right)\orbr{\begin{cases}3x-1=1\\2x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=0\end{cases}}}\)

\(\Rightarrow x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}\)

Vì \(\left(3x-6\right)^{20}=\left(3x-6\right)^{10}\) nên \(3x-6=-1\) hoặc \(3x-6=0\) hoặc \(3x-6=1\)

(+) \(3x-6=-1=>x=\frac{5}{3}\) 

(+)\(3x-6=0=>x=2\) 

(+) \(3x-6=1=>x=\frac{7}{3}\)

10(x-20)=10

    x - 20 = 10-10

    x - 20 = 0

    x      = 0 + 20

     x     = 20

597 - 3x = 9.2

597 -3x = 18

      3x   = 597 - 18

      3x  = 579

       x   = 579 : 3

       x   = 193

10 + 2x = 1024 : 64

10 + 2x = 16

       2x  = 16 - 10

       2x  = 6

         x  = 6 : 2

         x  = 3

( 3x - 10 ) : 10 = 20

3x - 10           = 20 : 10 

3x - 10           = 2

3x                 = 2 + 10

3x                 = 12

x                   = 12 : 3

x                   = 4

3x-10=200

3x=210

x=70