Question

[Đào Thị Thùy Dương]

tìm x sao cho:

a) 1-2x<7

b) (x-1)*(x-2)>0

c) (x-2)^2 * (x+1) * (x-4)<0

d) 5/x<1

Answer 1

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

a) \(1-2x< 7\)

\(\Rightarrow2x>-6\)

\(\Rightarrow x>-3\)

Vậy \(x>-3\)

b) \(\left(x-1\right)\left(x-2\right)>0\)

\(\Rightarrow\hept{\begin{cases}x-1>0\\x-2>0\end{cases}}\)

hoặc \(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x>1\\x>2\end{cases}}\)

hoặc \(\hept{\begin{cases}x< 1\\x< 2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>2\\x< 1\end{cases}}\)

Vậy \(x>2\)hoặc \(x< 1\)

c) \(\left(x-2\right)^2\left(x-1\right)\left(x-4\right)< 0\left(1\right)\)

Vì \(\left(x-2\right)^2\ge0\forall x\)nên từ \(\left(1\right)\): \(\Rightarrow\left(x-2\right)^2>0\)

\(\Rightarrow\orbr{\begin{cases}x-2>0\\x-2< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>2\\x< 2\end{cases}}\)

Với : 

• \(x>2\)

\(\Rightarrow\hept{\begin{cases}x+1>3>0\\x-4>-2\end{cases}}\)

Nếu \(-2< x-4< 0\)thì \(2< x< 4\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)(Thỏa mãn)

Nếu \(x-4\ge0\)thì \(x\ge4\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)\ge0\)(Không thỏa mãn)

• \(x< 2\)

\(\Rightarrow\hept{\begin{cases}x+1< 3\\x-4< -2\end{cases}}\)

Nếu \(0< x+1< 3\)thì \(-1< x< 2\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)(Thỏa mãn)

Nếu \(x+1\le0\)thì \(x\le-1\)

\(\Rightarrow\left(x-2\right)^2\left(x+1\right)\left(x-4\right)\ge0\)(Không thỏa mãn)

Vậy \(2< x< 4\)hoặc \(-1< x< 2\)

d) \(\frac{5}{x}< 1\)

\(\Rightarrow5< x\)

Vậy \(x>5\)