\(\dfrac{15x-2}{4}-\dfrac{x^2+1}{3}>\dfrac{x\left(1-2x\right)}{6}+\dfrac{x-3}{2}\\ \Leftrightarrow3\left(15x-2\right)-4\left(x^2+1\right)>2x\left(1-2x\right)+6\left(x-3\right)\\ \Leftrightarrow45x-6-4x^2-4>2x-4x^2+6x-18\\ \Leftrightarrow45x-6x-2x>6+4-18\\ \Leftrightarrow37x>-8\\ \Leftrightarrow x>-\dfrac{8}{37}\)

\(\dfrac{3\left(15x-2\right)}{12}-\dfrac{4\left(x^2+1\right)}{12}>\dfrac{2x\left(1-2x\right)}{12}+\dfrac{6\left(x-3\right)}{12}\)

\(45x-6-\left(4x^2+4\right)>2x-4x^2+6x-18\)

\(45x-4x^2+4x^2-2x-6x>6+4-18\)

\(37x>-8\)

\(x>\dfrac{-8}{37}\)

a: =>5(2-x)<3(3-2x)

=>10-5x<9-6x

=>x<-1

b: =>2/9x+5/3>=1/5x-1/5+1/3x

=>2/9x+5/3>=8/15x-1/5

=>-14/45x>=-28/15

=>x<=6

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{3-x}\right)}{2\left(x-1\right)}>x-\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)

- TH1: Với \(x>1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.

b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).

a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)

\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)

\(\Leftrightarrow4-2x-6x-12\le3x-51\)

\(\Leftrightarrow-11x\le-43\)

\(\Leftrightarrow x\ge\dfrac{43}{11}\)

Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }

b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)

\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)

\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)

\(\Leftrightarrow0x\le-10\) (vô lý)

Vậy \(S=\varnothing\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

a: =>x^2-8x+16<x^2-8x

=>16<0(loại)

b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)

=>x+1/2>=5/3x-1

=>-2/3x>=-3/2

=>x<=3/2:2/3=9/4

c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)

=>21-3x<=8x-16

=>-11x<=-37

=>x>=37/11

a) 2x - 3 > 3(x - 2)

⇔ 2x - 3 > 3x - 6

⇔ 2x - 3x > -6 + 3

⇔ -x > -3

⇔ x < 3

Vậy S = {x | x < 3}

b) (12x + 1)/12 ≤ (9x + 1)/3 - (8x + 1)/4

⇔ 12x + 1 ≤ 4(9x + 1) - 3(8x + 1)

⇔ 12x + 1 ≤ 36x + 4 - 24x - 3

⇔ 12x - 36x + 24x ≤ 4 - 3 - 1

⇔ 0x ≤ 0 (luôn đúng với mọi x)

Vậy S = R

a: =>2x-3>3x-6

=>-x>-3

=>x<3

b: =>12x+1<=36x+4-24x-3

=>12x+1<=12x+1

=>0x<=0(luôn đúng)

a) \(2x-3>3\left(x-2\right)\)

\(\Leftrightarrow2x-3>3x-6\)

\(\Leftrightarrow2x-3x>-6+3\)

\(\Leftrightarrow-x>-3\)

\(\Leftrightarrow x< 3\)

Vậy bất phương trình có nghiệm là \(x< 3\)

b) \(\dfrac{12x+1}{12}\le\dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow\dfrac{12x+1}{12}\le\dfrac{\left(9x+1\right).4}{3.4}-\dfrac{\left(8x+1\right)3}{4.3}\)

\(\Leftrightarrow12x+1\le36x+4-24x-3\)

\(\Leftrightarrow12x-36x+24x\le4-3-1\)

\(\Leftrightarrow0x\le0\)

Vậy bất phương trình vô nghiệm