a) x² - 7x + 16
= (x² - 2x\(\frac{7}{2}\)+ \(\frac{49}{4}\)) + \(\frac{15}{4}\)
= (x - \(\frac{7}{2}\))² + \(\frac{15}{4}\)> 0
b) 3x² - 3x + 1
= [\(\left(\sqrt{3x^2}\right)^2\)- 2.\(\sqrt{3x^2}\).\(\frac{\sqrt{3}}{2}\)+ \(\frac{3}{4}\)] + \(\frac{1}{4}\)
= (\(\sqrt{3x^2}\)- \(\frac{\sqrt{3}}{2}\))² + \(\frac{1}{4}\)> 0
c) -x² + 3x - 5
= -(x² - 3x + 5)
= -(x² - 2x\(\frac{3}{2}\)+ \(\frac{9}{4}\)+\(\frac{11}{4}\))
= -[(x - \(\frac{3}{2}\))² + \(\frac{11}{4}\)] < 0
d) Câu này sai đề rồi bạn ơi

giai giup minh dc ko minh dang can gap

\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0\)

\(x^2+3x+6=x^2+3x+\frac{9}{4}+\frac{15}{4}=\left(x+\frac{3}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

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

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

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

a) Vì (x+1)(x-2)<0 nên x+1 và x-2 trái dâu. Mà x+1> x-2 nên x+1>0  => x > -1   ( x thuộc Q)

       x-2<0         x < 2

Vậy -1< x < 2 ( x thuộc Q)