a) TH1:
\(\left\{{}\begin{matrix}x< 0\\8-x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x< 8\end{matrix}\right.\) \(\Rightarrow x< 0\)
TH2:
\(\left\{{}\begin{matrix}x>0\\8-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>0\\x>8\end{matrix}\right.\)\(\Rightarrow x>8\)
Vậy \(\left[{}\begin{matrix}x< 0\\x>8\end{matrix}\right.\)
b) TH1:
\(\left\{{}\begin{matrix}2-x>0\\x+3>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 2\\x>-3\end{matrix}\right.\)\(\Rightarrow-3< x< 2\)
TH2:
\(\left\{{}\begin{matrix}2-x< 0\\x+3< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)(vô nghiệm)
Vậy \(-3< x< 2\)
c) TH1:
\(\left\{{}\begin{matrix}2x-4>0\\5-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x>4\\x>5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>5\end{matrix}\right.\Rightarrow x>5\)
TH2:
\(\left\{{}\begin{matrix}2x-4< 0\\5-x>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x< 4\\x< 5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< 5\end{matrix}\right.\Rightarrow x< 2\)
Vậy \(\left[{}\begin{matrix}x>5\\x< 2\end{matrix}\right.\)
a, \(x\left(8-x\right)< 0\\ 8x-x^2< 0\)
Có x2 ≥ 0 ∀ x (1)
\(\Rightarrow\) - x2 ≤ 0 ∀ x
Mà 8x - x2 < 0
\(\Rightarrow\) 8x < x2 (2)
Thay (1) vào (2) \(\Rightarrow\) 8x < 0
\(\Rightarrow\) x < 0
Vậy x < 0
b, \(\left(2-x\right)\left(x+3\right)>0\\ 2x+6-x^2-3x>0\\ \Rightarrow\left(6-x\right)-x^2>0\)
Có x2 ≥ 0 ∀ x (1)
⇒ - x2 ≤ 0 ∀ x
Mà (6 - x) - x2 > 0
\(\Rightarrow6-x>x^2\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow6-x>0\\ \Rightarrow x< 0\)
Vậy x < 0
c, \(\left(2x-4\right)\left(5-x\right)< 0\\ 10x-2x^2-20+4x< 0\\ \Rightarrow\left(-20+14x\right)-2x^2< 0\)
Có x2 ≥ 0 ∀ x
⇒ 2x2 ≥ 0 ∀ x (1)
⇒ - 2x2 ≤ 0 ∀ x
Mà (-20 + 14x) - 2x2 < 0
\(\left(-20+14x\right)< 2x^2\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow-20+14x< 0\\ \Rightarrow14x< 20\\ \Rightarrow x< \frac{10}{7}\)
Vậy \(x< \frac{10}{7}\)
