3\(x^2\) - 4\(x\) - 4 = 0

3(\(x^2\) - 2. \(\dfrac{2}{3}\)\(x\) + \(\dfrac{4}{9}\)) - \(\dfrac{16}{3}\) = 0

3.(\(x-\dfrac{2}{3}\))² = \(\dfrac{16}{3}\)

   (\(x-\dfrac{2}{3}\))² = \(\dfrac{16}{9}\) 

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

  \(\left[{}\begin{matrix}x=\dfrac{4}{3}+\dfrac{2}{3}\\x=-\dfrac{4}{3}+\dfrac{2}{3}\end{matrix}\right.\)

  \(\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

S = { -\(\dfrac{2}{3}\); 2}

3x² - 4x - 4 = 0

⇔ 3x² - 6x + 2x - 4 = 0

⇔ (3x² - 6x) + (2x - 4) = 0

⇔ 3x(x - 2) + 2(x - 2) = 0

⇔ (x - 2)(3x + 2) = 0

⇔ x - 2 = 0 hoặc 3x + 2 = 0

*) x - 2 = 0

⇔ x = 2

*) 3x + 2 = 0

⇔ 3x = -2

⇔ x = -2/3

Vậy S = {-2/3; 2}

khó quá ?????

mik chưa học đến lớp 8 nên ko biết

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

- Với \(x=0\) không phải nghiệm

- Với \(x\ne0\) chia 2 vế cho \(x^2\)

\(\Rightarrow3\left(x^2+\dfrac{4}{x^2}\right)-2\left(x+\dfrac{2}{x}\right)-52=0\)

Đặt \(x+\dfrac{2}{x}=t\Rightarrow t^2=x^2+\dfrac{4}{x^2}+4\Rightarrow x^2+\dfrac{4}{x^2}=t^2-4\)

Pt trở thành:

\(3\left(t^2-4\right)-2t-52=0\)

\(\Leftrightarrow3t^2-2t-64=0\)

Nghiệm của pt này xấu quá

\(4x^2-4x-5\left|2x-1\right|-5=0\)

\(\Leftrightarrow-5\left|2x-1\right|=5-4x^2+4x\)

\(\Leftrightarrow\left|2x-1\right|=\frac{-4x^2+4x+5}{-5}\)

\(\Leftrightarrow\left|2x-1\right|=\frac{4x\left(x-1\right)}{5}-1\)

TH1 : \(2x-1=\frac{4x\left(x-1\right)}{5}-1\Leftrightarrow2x=\frac{4x\left(x-1\right)}{5}\)

\(\Leftrightarrow10x=4x^2-4x\Leftrightarrow14x-4x^2=0\)

\(\Leftrightarrow-2x\left(2x-7\right)=0\Leftrightarrow x=0;x=\frac{7}{2}\)

TH2 : \(2x-1=-\left(\frac{4x\left(x-1\right)}{5}-1\right)\Leftrightarrow2x-1=-\frac{4x\left(x-2\right)}{5}+1\)

\(\Leftrightarrow2x-2=-\frac{4x\left(x-2\right)}{5}\Leftrightarrow10x-10=-4x^2+8x\)

\(\Leftrightarrow2x-10+4x^2=0\Leftrightarrow2\left(2x^2+x-5\ne0\right)=0\)tự chứng minh 

Vậy tập nghiệm của phương trình là S = { 0 ; 7/2 }

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)