1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

a) \(\dfrac{15-6x}{3}>5\Leftrightarrow15-6x>15\)

\(\Leftrightarrow-6x>0\Leftrightarrow x< 0\) (vì \(-6< 0\))

\(S=\left\{x|x< 0\right\}\)

b) \(\dfrac{8-11x}{4}< 13\Leftrightarrow8-11x< 52\)

\(\Leftrightarrow-11x< -44\Leftrightarrow x>4\) (vì \(-11< 0\))

\(S=\left\{x|x>4\right\}\)

c) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

\(\Leftrightarrow8x+3x+1>5x-2x+6\)

\(\Leftrightarrow8x+3x-5x+2x>6-1\)

\(\Leftrightarrow8x>5\)

\(\Leftrightarrow x>\dfrac{5}{8}\) (vì \(8>0\))

\(S=\left\{x|x>\dfrac{5}{8}\right\}\)

d) \(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>-6\)

\(\Leftrightarrow-3x>-6\)

\(\Leftrightarrow x< 2\) (vì \(-3< 0\))

\(S=\left\{x|x< 2\right\}\)

a) \(\dfrac{15-6x}{3}>5\) <=> \(15-6x>15\) <=> \(6x< 0\) <=> \(x< 0\)

b) \(\dfrac{8-11x}{4}< 13\) <=> \(8-11x< 52\) <=> \(11x>-44\)<=> \(x>-4\)

c) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

<=> 8x + 3x + 3 - 5x + 2x - 6 > 0

<=> 8x  > 3

<=> x > 3/8

d) 2x(6x - 1) > (3x - 2)(4x + 3)

<=> 12x² - 2x > 12x² + x - 6

<=> 12x² - 2x - 12x² - x > -6

<=> -3x > -6

<=> x < 2

A, 3X+6>0

(=)3X>-6

(=)X>-2

VẬY ...

B,10-2X≥-4

(=)-2X≥-4-10

(=)-2X≥-14

(=)X≤7

VẬY....

C,

(=)

(=) -15X+10>-3+3X

(=)-15X-3X>-3-10

(=)-18X>-13

(=)X<

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

\(f\left(x\right)=\dfrac{\left(3x-4\right)\left(2x-3\right)}{\left(x^2-5x+6\right)\left(5-x\right)}>0\)

\(\Leftrightarrow\dfrac{\left(3x-4\right)\left(2x-3\right)}{\left(x-2\right)\left(x-3\right)\left(5-x\right)}>0\)

Bảng xét dấu:

Từ bảng xét dấu ta thấy nghiệm của BPT là: \(\left[{}\begin{matrix}x< 5\\\dfrac{3}{2}< x< 2\\3< x< 5\end{matrix}\right.\)