a)

 Ta có

 \(\left(x-1\right)^3-\left(x-1\right)^3-\left(6x-1\right)=-10\)

\(\Leftrightarrow-6x+1=-10\)

\(\Leftrightarrow-6x=-11\)

\(\Leftrightarrow x=\frac{11}{6}\)

  Vậy \(x=\frac{11}{6}\)

a) ( x - 1 )³ - ( x - 1 )³ - ( 6x - 1 ) = -10

<=> -( 6x - 1 ) = -10

<=> -6x + 1 = -10

<=> -6x = -11

<=> x = 11/6

b) ( 2x - 1 )² + ( 2x - 1 )( 2x - 3 ) - ( 2x + 3 )² + ( 2x + 3 )( -3x ) - 24 = 4

<=> 4x² - 4x + 1 + 4x² - 8x + 3 - ( 4x² + 12x + 9 ) - 6x² - 9x - 24 = 4

<=> 4x² - 4x + 1 + 4x² - 8x + 3 - 4x² - 12x - 9 - 6x² - 9x - 24 = 4

<=> -2x² - 33x - 29 - 4 = 0

<=> -2x² - 33x - 33 = 0 ( muốn kết quả thì ib còn mình để là vô nghiệm vì nó có nghiệm vô tỉ )

=> Vô nghiệm 

a) \(\left(x-1\right)^3-\left(x-1\right)^3-\left(6x-1\right)=-10\)

\(\Leftrightarrow6x-1=10\)

\(\Leftrightarrow6x=11\)

\(\Rightarrow x=\frac{11}{6}\)

b) \(\left(2x-1\right)^2+\left(2x-1\right)\left(2x-3\right)-\left(2x+3\right)^2+\left(2x+3\right).\left(-3x\right)-24=4\)

\(\Leftrightarrow\left(2x-1\right)\left(4x-4\right)-\left(2x+3\right)\left(5x+3\right)-28=0\)

\(\Leftrightarrow8x^2-12x+4-10x^2-21x-9-28=0\)

\(\Leftrightarrow2x^2+33x+33=0\)

\(\Leftrightarrow x^2+\frac{33}{2}x+\frac{33}{2}=0\)

\(\Leftrightarrow\left(x^2+\frac{33}{2}x+\frac{1089}{16}\right)-\frac{825}{16}=0\)

\(\Leftrightarrow\left(x+\frac{33}{4}\right)^2-\left(\frac{5\sqrt{33}}{4}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{33-5\sqrt{33}}{4}\right)\left(x+\frac{33+5\sqrt{33}}{4}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{5\sqrt{33}-33}{4}\\x=-\frac{33+5\sqrt{33}}{4}\end{cases}}\)

\(x^3-2x=-x^2+2\)

\(\Leftrightarrow x^3+x^2-2x-2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Ta có: \(x^3-2x=-x^2+2\)

    \(\Leftrightarrow\left(x^3+x^2\right)-\left(2x+2\right)=0\)

    \(\Leftrightarrow x^2.\left(x+1\right)-2.\left(x+1\right)=0\)

    5\(\Leftrightarrow\left(x+1\right).\left(x^2-2\right)=0\)

    \(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{2}\end{cases}}\)

Vậy \(S=\left\{-\sqrt{2};-1;\sqrt{2}\right\}\) 

\(x^3-2x=-x^2+2\)

\(x^3+x^2-2x-2=0\)

\(x^2\left(x+1\right)-2\left(x+1\right)=0\)

\(\left(x+1\right)\left(x^2-2\right)=0\)

\(Th1:x+1=0\Leftrightarrow x=-1\)

\(Th2:x^2-2=0\Leftrightarrow x^2=2\Leftrightarrow x=\pm\sqrt{2}\)

https://olm.vn/hoi-dap/detail/233628753470.html

vì | a | = | -a | \(\forall\)a

| 2x - 1 | = | - ( 2x - 1 ) | = | 1 - 2x | 

nên | 2x - 1 | + | 1 - 2x | = 10

hay 2 . | 2x - 1 | = 10

| 2x - 1 | = 10 : 2 = 5

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=5+1=6\\2x=-5+1=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=6:2=3\\x=-4:2=-2\end{cases}}\)

Vậy x = { 3 ; -2 }

ta mở lòng từ bi cho con 1 tát

Bạn viết dấu giá trị tuyệt đối kiểu gì vậy ?

x/6 = y/5 \(\Rightarrow\)x/12 = y/10

x/4 = z/7 \(\Rightarrow\)x/12 = x/21

Vì x/12 =y/10 ; x/12= z/21

\(\Rightarrow\)x/12 = y/10 = z/21

Ta có : x/12 = 2x/24

y/10 = 3y/30

Áp dụng tính chất của dãy tỉ sô bằng nhau ta có :

2x/24 = 3y/30= z/21 = 2x -3y + z / 24 - 30 + 21 = 75/ 15= 5

Ta có : x/12 = 5 \(\Rightarrow\)x = 5 x 12 = 60

y/10 =5 \(\Rightarrow\)y = 5 x 10 = 50

z/21 = 5 \(\Rightarrow\)z = 21 x 5 = 105

Vậy x = 60

 y  = 50

 z = 105

\(\frac{x-1}{-3}\)=\(\frac{2x+5}{5}\)\(\Rightarrow\)5(x-1)=-3(2x+5)

                                           \(\rightarrow\)5x-5=-6x-15

                                          \(\rightarrow\)5x+6x=5-15

                                         \(\rightarrow\)11x=-10\(\Rightarrow\)x=\(\frac{-10}{11}\)

vậy.........................

\(a,\left|2x-1\right|>0\Rightarrow x\in z\)

\(b,\left|2x-1\right|< 0\)

x ko có số nào cả