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

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

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

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

\(37x>-8\)

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

ĐKXĐ: \(1< x< 9\)

Đặt \(\left\{{}\begin{matrix}\sqrt{9-x}=a\\\sqrt{x-1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a;b>0\\a^2+b^2=8\end{matrix}\right.\) \(\Rightarrow\left(a+b\right)^2\le16\Rightarrow a+b\le4\)

\(BPT\Leftrightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}\ge3\) (1)

Đặt \(P=\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}-3\)

\(P=a+b-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-3\le a+b-\dfrac{4}{a+b}-3\)

\(P\le\dfrac{\left(a+b\right)^2-3\left(a+b\right)-4}{a+b}=\dfrac{\left(a+b+1\right)\left(a+b-4\right)}{a+b}\le0\)

\(\Rightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}\le3\) (2)

(1); (2) \(\Rightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}=3\)

Dấu "=" xảy ra khi và chỉ khi: \(a=b=2\Leftrightarrow x=5\)

Vậy BPT đã cho có nghiệm duy nhất \(x=5\)

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

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=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)

a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)

\(\Leftrightarrow6x+2-20+8x>8x-6-6\)

\(\Leftrightarrow14x-18-8x+12>0\)

\(\Leftrightarrow6x-6>0\)

\(\Leftrightarrow6x>6\)

hay x>1

Vậy: S={x|x>1}

b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)

\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)

\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)

\(\Leftrightarrow-1< 0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}