15 người                                    

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

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)

16 + ( - 6) = 10

b) 

14 + ( - 6 ) = 8

c)

( - 8 ) +12 =4

Áp dụng tc của dãy tỉ số bằng nhau ta có

\(\frac{x_1-1}{100}=\frac{x_2-2}{99}=.....=\frac{x_{100}-100}{1}=\frac{\left(x_1+x_2+....+x_{100}\right)-\left(1+2+...+100\right)}{1+2+3+....+100}\)

\(=\frac{15150-2525}{2525}=\frac{12625}{2525}=5\)

\(\Rightarrow\begin{cases}x_1=100.5+1\\x_2=99.5+2\\......\\x_{10}=1.5+100\end{cases}\)

\(\Rightarrow\begin{cases}x_1=501\\x_2=497\\...\\x_{100}=105\end{cases}\)

Mính xin lỗi

\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right)......\left(1^2-2016^2\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)

\(A=\frac{\left(-1\right)3\left(-2\right)4..........\left(-2015\right)2017}{\left(2.3....2016\right)\left(2.3...2016\right)}\)

\(A=\frac{\left[\left(-1\right)\left(-2\right)....\left(-2015\right)\right]\left(3.4.5...2017\right)}{\left(2.3.4.....2016\right)\left(2.3.4...2017\right)}\)

\(A=\frac{\left(-1\right)2017}{2016.2}=-\frac{2017}{4032}< -\frac{2016}{4032}=-\frac{1}{2}\)

=> A< - 1/2

3x/2-1/3=2/5+x

3x/2-x=2/5+1/3

1/2x=11/15

x=11/15:1/2

x=22/15

đúng-**** cho mk nha!!!

diện tích của tam giác ABC là

1/2*6*(3+2)=15 \(cm^2\)

diện tích tam giác BDC là

1/2*2*6=6 \(cm^2\)

diện tích phần tô đậm là

15-6=9 \(cm^2\)

đáp số   9 \(cm^2\)

Ta có

\(\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\\x+2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=-2\end{array}\right.\)

Vậy x=1 ; x=2 ; x= - 2

Ta có

\(\left(x+1\right)\left(x-1\right)=x^2-1\)

\(\Rightarrow x^2-1\le0\)

\(\Rightarrow x^2\le1\)

Mà \(x^2\ge0\) với mọi x

\(\Rightarrow\left[\begin{array}{nghiempt}x^2=0\\x^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\end{array}\right.\)

Vậy x=0 ; x=1 ; x=-1

\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right)..........\left(1^2-2016^2\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)\left(3\right)\left(-2\right)\left(4\right)....\left(-2015\right)\left(2017\right)}{\left(2.3....2016\right)\left(2.3...2016\right)}\)

\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right).....\left(-2015\right)\right]\left(3.4.5...2017\right)}{\left(2.3.....2016\right)\left(2.3.4....2016\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)2017}{2016}=-\frac{2017}{2016}< \frac{1}{2}\)

=> A<1/2