1

b;

B=1+ (7-5) + (11-9) + ...+(101-99)

B=1+2+2+..+2

B=1+25.2=51

2.

a.

ĐK : x+2 >=0 => x>=-2

\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)

Vậy x=-2

ai bt tự làm

ngu tự chịu

       A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101

=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4

=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)

=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101

=> 4A = 99*100*101*102

=> 4A = 101989800

=>   A = 25497450

bạn vào câu hỏi tương tự tham khảo nhé!

Vậy mà cũng gọi là trả lời 

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)

\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)

\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)

....

\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)

=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)

=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)

=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)

=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)

=\(\frac{5}{2}\cdot\frac{100}{101}\)

\(=\frac{250}{101}\)