a, 7\(x\).(2\(x\) + 10) =0

    \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\) {-5; 0}

b, -9\(x\) : (2\(x\) - 10) = 0

    9\(x\)                   = 0 

     \(x\)                    = 0 

c, (4 - \(x\)).(\(x\) + 3)  = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

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

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

     \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\in\) {-2023; 2024}

a)\(=>2x=-10=>x=-5\)

b)\(=>-2x=-5=>x=\dfrac{-5}{-2}=\dfrac{5}{2}\)

c)\(4-x=0=>x=4-0=4\)

d)\(=>2x=-1=>x=-\dfrac{1}{2}\)

e)\(=>x^2=-2\)=> x ko tồn tại

f)\(=>x\left(2+1\right)=0=>3x=0=>x=0\)

a) 2x+8≤ 0 

⇔2x≤-8

⇔x≤-4

b) 4x-7 ≥ 2x -5

⇔2x-12 ≥ 0

⇔2x≥12

⇔x≥6

c) (2x-8)(15-3x)>0

TH1: 2x-8>0 ⇒x>4

        15-3x>0⇒x<5 

TH2:  2x-8<0 ⇒x<4

        15-3x<0⇒x>5 (vô lí)

vậy 4<x<5

3x(x - 10) = x - 10

(x - 10)(3x - 1) = 0

Th1:

x - 10 = 0

x = 10

TH2:

3x - 1 = 0

3x = 1

x = 1/3

Vậy x = 10 hoặc x = 1/3

x(x + 7) - (4x + 28) = 0

x(x + 7) - 4(x + 7) = 0

(x + 7)(x - 4) = 0

Th1:

x + 7 = 0

x = - 7

Th2:

x - 4 = 0

x = 4

Vậy x = - 7 hoặc x = 4

x(x - 4) = 2x - 8

x(x - 4) - 2(x - 4) = 0

(x - 2)(x - 4) = 0

Th1:

x - 2 = 0

x = 2

Th2:

x - 4 = 0

x = 4 

Vậy x = 2 hoặc x = 4

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

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

4x(x - 1) = 0

Th1:

x = 0

Th2:

x - 1 = 0

x = 1

Vậy x = 0 hoặc x = 1

a)

\(3x\left(x-10\right)=x-10\)

\(\Rightarrow3x\left(x-10\right)-\left(x-10\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(x-10\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-1=0\\x-10=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=10\end{array}\right.\)

b)

\(x\left(x+7\right)-\left(4x+28\right)=0\)

\(\Rightarrow x\left(x+7\right)-4\left(x+7\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x+7\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=-7\end{array}\right.\)

c)

\(x\left(x-4\right)=2x-8\)

\(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)

d)

\(\left(2x+3\right)\left(x-1\right)+\left(2x+3\right)\left(x-1\right)=0\)

\(\Rightarrow2\left(2x+3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x+3=0\\x-1=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=1\end{array}\right.\)

a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

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

<=> | 2x - 10 | = 2x - 10

<=> 2x -10 ≥ 0

<=> 2x ≥ 10

<=> x ≥ 5

mà x thuộc Z

=> x thuộc Ơ 5;6;7;8;9;...Ư

a,  7\(x\).(2\(x\) + 10) = 0

        \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-10:2\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(x\in\){-5; 0}

b, - 9\(x\) : (2\(x\) - 10) = 0

      - 9\(x\) = 0

           \(x\) = 0

c, (4 - \(x\)).(\(x\) + 3) = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

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

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

    \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-2023; 2024}

Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0