\(\Leftrightarrow\dfrac{x}{a}-x>4-a-\dfrac{3}{a}\)

\(\Leftrightarrow x\left(\dfrac{1}{a}-1\right)>\dfrac{4a-a^2-3}{a}\)

- Nếu \(\dfrac{1}{a}-1>0\Leftrightarrow0< a< 1\)

\(\Rightarrow x>\dfrac{4a-a^2-3}{a\left(\dfrac{1}{a}-1\right)}\Leftrightarrow x>a-3\)

- Nếu \(\dfrac{1}{a}-1< 0\Leftrightarrow\left[{}\begin{matrix}a< 0\\a>1\end{matrix}\right.\)

\(\Rightarrow x< \dfrac{4a-a^2-3}{a\left(\dfrac{1}{a}-a\right)}\Leftrightarrow x< a-3\)

Điều kiện xác định: a ≠ 0.

Ta có:

⇔ x( a + 2 ) > 1/a    ( 1 )

+ Nếu a > - 2,a ≠ 0 thì nghiệm của bất phương trình là

+ Nếu a < - 2 thì nghiệm của bất phương trình là

+ Nếu x = - 2 thì ( 1 ) có dạng 0x > - 1/2 luôn đúng với ∀ x ∈ R

ĐKXĐ: x\(\ne3,x\ne-3\) 

\(\Rightarrow\left(x-a\right)\left(a-3\right)+\left(x+3\right)\left(a+3\right)=-6a\) 

\(\Leftrightarrow xa-3x-a^2+3a+ax+3x+3a+3=-6a\)

\(\Leftrightarrow2ax-a^2+12a+3=0\) \(\Leftrightarrow2ax=a^2-12a-3\Leftrightarrow x=\dfrac{a^2}{2}-6a-\dfrac{3}{2}\)(TM)

Vậy...

\(\dfrac{x+1}{a}+ax>\dfrac{x+2}{a}-2x\)

⇔ \(\dfrac{x}{a}+\dfrac{1}{a}+ax>\dfrac{x}{a}+\dfrac{2}{a}-2x\) ( a # 0)

⇔ \(ax+2x>\dfrac{2}{a}-\dfrac{1}{a}\)

⇔ \(x\left(a+2\right)>\dfrac{1}{a}\) ( 1)

+) Với : a = -2 , ta có :

( 1) ⇔ 0x > \(\dfrac{-1}{2}\) ( Luôn đúng )

+) Với : a > -2 , ta có :

( 1) ⇔x > \(\dfrac{1}{a\left(a+2\right)}\)

+) Với : a < - 2 , ta có :

⇔ x < \(\dfrac{1}{a\left(a+2\right)}\)

KL...

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

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

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, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...