|2x-10|+2x-10=0

=> |2x-10| = 0 - (2x+10)

=> |2x-10| = - (2x+10)

=> 2x-10 < = 0

=> 2x < = 10

=> x < = 10 : 2

=> x < = 5

Vậy x < = 5

Tk mk nha

\(\left|2x-10\right|+2x-10=0\)

\(\Leftrightarrow\left|2x-10\right|=-\left(2x-10\right)\)

\(\Leftrightarrow2x-10\le0\)

\(\Leftrightarrow2x\le10\)

\(\Leftrightarrow x\le5\)

nhìn tui đây

hghghghghghghghghgghhhhgghgh

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow2x+3⋮2x-1\)

\(\Leftrightarrow2x-1\inƯ\left(4\right)\)

\(\Leftrightarrow2x-1\in\left\{-1;1\right\}\)

hay \(x\in\left\{0;1\right\}\)

\(\left|x+6\right|-9=2x\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+6-9=2x\\x-6+9=2x\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-2x=-6+9\\x-2x=6-9\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}-x=3\\-x=-3\end{array}\right.\)

Vậy \(x=-3\)

\(\left(2x-3\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\) (Thêm KL cuối dòng: Vậy \(x\in\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\))

(2x-3)x(x-1/2)=0

Đặt từng nhân tử bằng không và giải cho x:

2x - 3 = 0

2x = 3

x = 3/2

x = 0

x - 1/2 = 0

x = 1/2

a:3 dư 3 nên a-3: hết cho 4 nên a-23 chia hết cho 4

a:9 dư 5 nên a-5 chia hết cho 9 nên a-23 chia hết cho 9

nên a-23 chia hết cho 9 và 4 mà UCLN(9;4)=1 nên a-23 chia hết cho 36

a chia 36 dư 23

k nha bạn