ĐKXĐ : x khác -1

\(\dfrac{x^2+2x+2}{x+1}\ge\dfrac{x^2+3x+4}{x+1}\\ \Leftrightarrow\dfrac{x^2+2x+2}{x+1}\ge\dfrac{x^2+2x+2}{x+1}+\dfrac{x+2}{x+1}\\ \Leftrightarrow\dfrac{x+2}{x+1}\le0\\ \Leftrightarrow x+2\ge0;x+1< 0\Leftrightarrow-1>x\ge-2\)

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

2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

a) \(\dfrac{2x+1}{x-2}=3\Rightarrow2x+1=3x-6\Rightarrow x=7\)

b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\Rightarrow4x-6=x+1\Rightarrow3x=7\Rightarrow x=\dfrac{7}{3}\)

a) \(\dfrac{2x+1}{x-2}=3\)

dkxd : x ≠ 2

MTC : x - 2

Quy đồng mẫu thức : 

⇒ \(\dfrac{2x+1}{x-2}=\dfrac{3\left(x-2\right)}{x-2}\)

Suy ra : 2x + 1 = 3(x - 2)

      \(\) \(\Leftrightarrow\) 2x + 1 = 3x - 6

       \(\Leftrightarrow\) 2x + 1 - 3x + 6 = 0

      \(\Leftrightarrow\) -1x + 7 = 0

     \(\Leftrightarrow\) -1x = -7

    \(\Leftrightarrow\) x = \(\dfrac{-7}{-1}=7\)

 Vậy S = \(\left\{7\right\}\)

b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\)

dkxd : x ≠ -1

MTC : 2(x + 1)

Quy đồng mẫu thức : 

⇒ \(\dfrac{2\left(2x-3\right)}{2\left(x+1\right)}=\dfrac{1\left(x+1\right)}{2\left(x+1\right)}\)

Suy ra : 2(2x - 3) = x + 1

        \(\Leftrightarrow\) 4x - 6 - x - 1 = 0

       \(\Leftrightarrow\) 3x - 7 = 0

       \(\Leftrightarrow\) 3x = 7

      \(\Leftrightarrow\) x = \(\dfrac{7}{3}\)

   Vậy S = \(\left\{\dfrac{7}{3}\right\}\)

 Chúc bạn học tốt

a: Ta có: \(\dfrac{2x+1}{x-2}=3\)

\(\Leftrightarrow3x-6=2x+1\)

hay x=7

b: Ta có: \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\)

\(\Leftrightarrow4x-6=x+1\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

1) \(ĐK:x\ne2\) 

Nếu \(x>2\) 

BPT ⇔ \(x^2-2x+5-\left(x-1\right)\left(x-2\right)\ge0\) ⇔ \(x^2-2x+5-\left(x^2-3x+3\right)\ge0\)

⇔\(x+2\ge0\) ⇔\(x\ge-2\) ⇒ Lấy \(x\ge2\)

Nếu \(x< 2\)

BPT ⇔\(\dfrac{-\left(x^2-2x+5\right)}{x-2}-x+1\ge0\) ⇔\(-x^2+2x-5-\left(x-1\right)\left(x-2\right)\ge0\)

⇔\(-x^2+2x-5-x^2+3x-2\ge0\)

⇔\(-2x^2+5x-7\ge0\)

⇔\(x^2-\dfrac{5}{2}x+\dfrac{7}{2}\le0\)

⇔\(\left(x-\dfrac{5}{4}\right)^2\le\dfrac{11}{4}\)

⇔\(\left[{}\begin{matrix}x-\dfrac{5}{4}\le\dfrac{11}{4}\\x-\dfrac{5}{4}\le\dfrac{-11}{4}\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x\le4\\x\le\dfrac{-3}{2}\end{matrix}\right.\) ⇔ \(x\le\dfrac{-3}{2}\) 

S= [2;+∞)U(-∞;\(\dfrac{-3}{2}\)]

2) \(ĐK:x\ne-1\) 

Nếu \(x>-1\) 

BPT ⇔ \(2x-3-2\left(x+1\right)< 0\) ⇔\(2x-3-2x-2< 0\)

 ⇔\(-5< 0\) ( luôn đúng với mọi \(x>-1\))

Nếu \(x< -1\)

BPT⇔\(\dfrac{-\left(2x-3\right)}{x+1}-2< 0\) ⇔\(-\left(2x-3\right)-2\left(x+1\right)< 0\) ⇔\(-4x+1< 0\) ⇔ \(x>\dfrac{-1}{4}\)

Vậy S=....

1.

ĐK: \(x\ne7;x\ne-1;x\ne3\)

\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)

TH1: \(x< -1\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\ge0\)

\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)

TH2: \(-1< x< 3\)

\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)

\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)

\(\Leftrightarrow-x^2+5x-22\ge0\)

\(\Rightarrow\) vô nghiệm

TH3: \(3< x< 7\)

Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)

\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)

TH4: \(x>7\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\le0\)

\(\Rightarrow\) vô nghiệm

Vậy ...

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