\(A=\frac{\left(\frac{2012-2}{2}+1\right)}{2}.\left(2+2012\right)+2014=1015056\)

Số số hạng : (2014 - 2) : 2 + 1 = 1007 (Số)

Tổng : (2 + 2014) x 1007 : 2 = 1015056

Chúc bạn hok tốt nha!

Ta có : 

\(xy-x+y-1=0\)

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\y=1\end{cases}}}\)

Vậy \(x=-1\) và \(y=1\)

Chúc bạn học tốt ~ 

 = x(y-1) +(y-1) =0 
(y-1)(x+1)=0
suy ra y-1=0 hoặc x+1=0 
suy ra : y=1 ;x=-1
vậy...

S= 338350

\(1^2+2^2+3^2+...+100^2\)

\(=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+....+100.\left(101-1\right)\)

\(=1.2-1+2.3-2+3.4-3+...+100.101-100\)

\(\left(1.2+2.3+3.4+...+100.101\right)-\left(1+2+3+...+100\right)\)

\(\Rightarrow3S=\left(1.2.3+2.3.3+3.4.3+...+100.101.3\right)-\frac{\left(1+100\right).100.3}{2}\)

\(\Rightarrow3S=\left(1.2.3+2.3.4-1.2.3+...+100.101.102-99.100.101\right)-5050.3\)

\(\Rightarrow3S=100.101.102-5050.3\)

\(\Rightarrow S=\frac{100.101.102}{3}-5050\)

Ta có : 

\(31^{11}< 32^{11}=\left(2^5\right)^{11}=2^{55}\)

\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}\)

Vì \(2^{55}< 2^{56}\) nên \(32^{11}< 16^{14}\)

\(\Rightarrow\)\(31^{11}< 32^{11}< 16^{16}< 17^{14}\)

\(\Rightarrow\)\(31^{11}< 17^{14}\)

Vậy \(31^{11}< 17^{14}\)

Chúc bạn học tốt ~ 

\(\frac{5}{3x}\)- \(\frac{2}{5x}\)= \(\frac{19}{10}\)

\(\frac{25}{15x}-\frac{6}{15x}=\frac{19}{10}\)

\(\frac{19}{15x}\)= \(\frac{19}{10}\)

\(\Rightarrow\)15x = 10

x = 10 : 15 = 10/15 = 2/3

Ta có : 220,1 < x + x + x + x + x < 225,2

        => 220,1 < 5x < 225,2

        => \(\frac{220,1}{5}< x< \frac{225,2}{5}\)

       => 44,02 < x < 45,04

Vậy 44,02 < x <45,04

TK nha! 

220,1 < 5x < 225,2
-> 44,02 < x < 45,04
-> x=45

x/2 - 1/x = 1/12 

6x/12 - 1/x = 1/12

1/x = 6x/12 - 1/12

1/x = 6x - 1 / 12

=> 1 . 12 = ( 6x - 1 ) . x

=> 12 = ( 6x - 1 ) . x 

thanks nha

\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5050\)

\(\left(100x\right)+\left(1+2+..+100\right)=5050\)

\(\left(100x\right)+5050=5050\)

\(100x=5050-5050\)

\(100x=0\)

\(x=0\div100\)

\(x=0\)

từ 1 đến 100 có 100 chữ số 

tổng của chúng sẽ là 

(100+1).100 : 2 = 5050 

mà : (x+1)+(x+2)+...+(x+100)=5050

thì x = 0

vậy x = 0