- Đặt \(f\left(x\right)=\dfrac{2x-3}{19+8x}\)

- Lập bảng xét dấu :

- Từ bảng xét dấu : - Để : \(f\left(x\right)< 0\)

\(\Leftrightarrow-\dfrac{19}{8}< x< \dfrac{3}{2}\)

Vậy ...

Ta có: \(\dfrac{2x-3}{8x+19}< 0\)

Trường hợp 1: \(\left\{{}\begin{matrix}2x-3>0\\8x+19< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{19}{8}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Trường hợp 2: \(\left\{{}\begin{matrix}2x-3< 0\\8x+19>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x>-\dfrac{19}{8}\end{matrix}\right.\Leftrightarrow-\dfrac{19}{8}< x< \dfrac{3}{2}\)

Vậy: S={x|\(-\dfrac{19}{8}< x< \dfrac{3}{2}\)}

\(\left|3-8x\right|< 19\Rightarrow\orbr{\begin{cases}3-8x< 19\\3-8x>-19\end{cases}\Rightarrow\orbr{\begin{cases}-8x< 16\\-8x>-22\end{cases}\Rightarrow\orbr{\begin{cases}x>-2\\x< \frac{22}{8}\end{cases}\Leftrightarrow}}-2< x< \frac{11}{4}}\)

\(A=\sqrt{16}\cdot\sqrt{25}+\sqrt{196}+\sqrt{19}\)

\(=4\cdot5+14+\sqrt{19}\)

\(=34+\sqrt{19}\)

\(A\left(x\right)=43x-\left(52x^2+34x^2-8x^4\right)-\left(8x^4+16x^3-42x^2+43x\right)+19\)

\(\Leftrightarrow A\left(x\right)=43x-86x^2+8x^4-16x^3+42x^2-43x+19\)

\(\Leftrightarrow A\left(x\right)=-16x^3-44x^2+19\)

Bậc là: 3

a) |3 - 8x| < 19

=> \(-19< 3-8x< 19\)

=> -22 < 8x < 16

=> -22 : 8 < x < 16 : 8

=> - 2,75  < x < 2

b) |x - 4| > 3

=> \(\orbr{\begin{cases}x-4>3\\-\left(x-4\right)< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>7\\-x+4< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>7\\x< 1\end{cases}}}}\)

B. -7

B

B