tách nhỏ câu hỏi ra nhé dài quá

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 ...

Các bài kia tương tự, chứ giải ra mệt lắm.

a) \(4x^2-x+1< 0\)

Tam thức f(x) = 4x² - x + 1 có hệ số a = 4 > 0 biệt thức ∆ = 1² – 4.4 < 0. Do đó f(x) > 0 ∀x ∈ R.

Bất phương trình 4x² - x + 1 < 0 vô nghiệm.

b) f(x) = - 3x² + x + 4 = 0

\(\Delta=1^2-4\left(-3\right).4=49\)

\(x_1=\dfrac{-1+\sqrt{49}}{-3}=-1\)

\(x_2=\dfrac{-1-\sqrt{49}}{-3.2}=\dfrac{4}{3}\)

- 3x² + x + 4 ≥ 0 <=> - 1 ≤ x ≤ .

d) -2;3

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. TH1:

\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

Vậy nghiệm của BPT:

\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)      \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)

a) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x-4\right)\left(x+4\right)\le10\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)\le10\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240\le10\)

\(\Leftrightarrow\left(5x^3-5x^3\right)-\left(30x^2-15x^2-15x^2\right)-\left(45x-15x\right)+5-240\le10\)

\(\Leftrightarrow30x-235\le10\)

\(\Leftrightarrow30x\le10+235\)

\(\Leftrightarrow30x\le245\)

\(\Leftrightarrow30x:30\le245:30\)

\(\Leftrightarrow x\le\dfrac{49}{6}\)

Vậy nghiệm của bất phương trình là: \(x\le\dfrac{49}{6}\)

b) \(\left(3x-2\right)\left(9x^2+6x+4\right)+27x\left(\dfrac{1}{3}-x\right)\left(\dfrac{1}{2}+x\right)\ge1\)

\(\Leftrightarrow27x^3-8+27x\left(\dfrac{1}{9}-x^2\right)\ge1\)

\(\Leftrightarrow27x^3-8+3x-27x^3\ge1\)

\(\Leftrightarrow\left(27x^3-27x^3\right)-8+3x\ge1\)

\(\Leftrightarrow-8+3x\ge1\)

\(\Leftrightarrow3x\ge1+8\)

\(\Leftrightarrow3x\ge9\)

\(\Leftrightarrow3x:3\ge9:3\)

\(\Leftrightarrow x\ge3\)

Vậy nghiệm của bất phương trình là \(x\ge3\)

a: =>5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15(x^2-16)<=10

=>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240<=10

=>30x-235<=10

=>30x<=245

=>x<=49/6

b: =>27x^3-8+27x(1/9-x^2)>=1

=>27x^3-8+3x-27x^3>=1

=>3x>=9

=>x>=3

ĐK: \(x\ge2\)

\(\dfrac{\sqrt{x^2+1}-\sqrt{x+1}}{x^2+\sqrt{3x-6}}\ge0\)

\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+1}\ge0\)

\(\Leftrightarrow\sqrt{x^2+1}\ge\sqrt{x+1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x^2+1\ge x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1\le x\le0\\x\ge1\end{matrix}\right.\)

Kết hợp điều kiện xác định ta được \(x\ge2\)