Question

[Jiyoen Phạm]

Tim x thuoc Q

a, (x+1)(x-2)<0

b, (x-2)(x+2/3)>0

Answer 1

[Isolde Moria]

\(\left(x+1\right)\left(x+2\right)< 0\)

Mà x+1 < x+2

\(\Rightarrow\begin{cases}x+1< 0\\x+2>0\end{cases}\)

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

\(\Rightarrow x\in\varnothing\)

b)

\(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

(+) Với \(\left(x-2\right);\left(x+\frac{2}{3}\right)\) cùng dương

\(\Rightarrow\begin{cases}x+2>0\\x+\frac{2}{3}>0\end{cases}\)

\(\Rightarrow\begin{cases}x>-2\\x>-\frac{2}{3}\end{cases}\)

=> x > - 2

(+) Với \(\left(x-2\right);\left(x+\frac{2}{3}\right)\) cùng âm

\(\Rightarrow\begin{cases}x+2< 0\\x+\frac{2}{3}< 0\end{cases}\)

\(\Rightarrow\begin{cases}x< -2\\x< -\frac{2}{3}\end{cases}\)

=> x < - 2

Vậy x>2 ; x< - 2

Answer 2

[Võ Đông Anh Tuấn]

a ) \(\left(x+1\right).\left(x-2\right)< 0\)

\(=x.\left(x-2\right)+1.\left(x-2\right)< 0\)

\(=x.\left(x-2\right)+\left(x-2\right)< 0\)

\(\Rightarrow x\in Z\)

\(\Rightarrow x>2\)

b ) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

\(=x.\left(x+\frac{2}{3}\right)-2.\left(x+\frac{2}{3}\right)\)

\(\Rightarrow\left(x+\frac{2}{3}\right)\in\)số nguyên

Nên \(x\in\) phấn số