( x - 6)(2x + 3) > 0 

(+) x - 6 > 0 và 2x + 3 > 0 

=> x > 6 và  x > -3/2

=> x > 6 

(+) x - 6 < 0  và 2x + 3 < 0 

=> x < 6 và x < -3/2

=> x< - 3/2 

Vậy x  < -3/2 và x >  0 với đk x thuộc Z thì ... 

Đug cho mình nha """

Sửa đề : a) Tìm GTNN A

a) \(A=\left|x-5\right|+3\)có : \(\left|x-5\right|\ge0\Rightarrow\left|x-5\right|+3\ge0\)

\(\Leftrightarrow A\ge3\)dấu "=" xảy ra khi : \(\left|x-5\right|=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Vậy GTNN A = 3 khi x = 5.

b) \(C=-\left|x+1\right|+5\)có : \(-\left|x+1\right|\le0\Rightarrow-\left|x+1\right|+5\le5\)

\(\Leftrightarrow C\le5\)dấu "=" xảy ra khi : \(-\left|x+1\right|=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy GTLN C = 5 khi x = -1.

\(D=5-\left|2x+3\right|\)có : \(-\left|2x+3\right|\le0\Rightarrow5-\left|2x+3\right|\le5\)

\(\Leftrightarrow D\le5\)dấu "=" xảy ra khi : \(-\left|2x+3\right|=0\Leftrightarrow2x+3=0\Leftrightarrow x=-\frac{3}{2}\)

Vậy GTLN D = 5 khi x = -3/2.

c) \(\left|x-3\right|+\left|y+1\right|=0\)có \(\left|x-3\right|\ge0;\left|y+1\right|\ge0\Rightarrow\left|x-3\right|+\left|y+1\right|\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x-3\right|=0\\\left|y+1\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-1\end{cases}}.\)

B=|2x+1|-4 

b, \(B=\left|2x+1\right|-4\) có : \(\left|2x+1\right|\ge0\)\(\Rightarrow\left|2x+1\right|-4\ge0\Rightarrow A\ge4\)

Dấu " = " xảy ra khi : \(\left|2x+1\right|=0\Rightarrow2x+1=0\Rightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

Vậy giá trị nhỏ nhất của B là 4 khi x = -1/2

a)\(\frac{1}{4}+\frac{1}{3}:2x=-5\)

   \(\frac{1}{3}:2x=-5-\frac{1}{4}\)

   \(\frac{1}{3}:2x=-\frac{21}{3}\)

   \(2x=\frac{1}{3}:\left(\frac{-21}{3}\right)\)

   \(2x=-\frac{1}{21}\)

   \(x=\frac{-1}{42}\)

b)\(\left(3x-\frac{1}{4}\right).\left(x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}3x=\frac{1}{4}\\x=-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{12}\\x=-\frac{1}{2}\end{array}\right.\)

c)\(\left(2x-5\right).\left(\frac{3}{2}x+9\right).\left(0,3x-12\right)=0\)

   \(\Rightarrow\left[\begin{array}{nghiempt}2x-5=0\\\frac{3}{2}x+9=0\\0,3x-12=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=5\\\frac{3}{2}x=-9\\0,3x=12\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-6\\x=40\end{array}\right.\)

a) 1/4 + 1/3 : 2x = -5

=> 1/3 : 2x = -5 - 1/4

=> 1/3 : 2x = -21/4

=> 2x = 1/3 : (-21/4) = -4/63

=> x = -4/63 : 2 = -2/63

a, 12 - (2\(x^2\) - 3) = 7

            2\(x^2\)  - 3  =  12  - 7

           2\(x^2\) - 3  = 5

           2\(x^2\)  = 8

             \(x^2\)   = 4

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

a) \(12-\left(2x^2-3\right)=7\\ 12-2x^2+3=7\\ 15-2x^2=7\\ 2x^2=15-7=8\\ x^2=8:2=4\\ x=\pm2\)

b) \(3x^2-12=2x^2+4\\ 3x^2-2x^2=12+4\\ x^2=16\\ x=\pm4\)

b, 3\(x^2\) - 12  = 2\(x^2\) + 4

    3\(x^2\) - 2\(x^2\) = 12 + 4

     \(x^2\) = 16

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

Cho hàm số y= f(x) =2x -3

a, Tính f(-3): f(0,5): f(0)

b, Tìm x biết f(x)=7

Trả lời:

a, f(-3): f(0,5): f(0)

=[2(-3)-3]:[2(0,5)]:(2.0-3)

=(-6-3):(-3)

=3

b, f(x)=y=7=2x-3

<=>2x-3=7

<=>2x=7+3

<=>2x=10

<=>x=5