a,

=(10-1)+(10^2   -  1)+...+(10^10  - 1)

=(10 + 10^2 + 10^3 +....+ 10^10) - 10

=10^2+10^3+10^4+....+10^10

=11111111100

b,

1/21+1/28 ko bằng 2/9

                     =>    \(3x-15=0\) 

                     =>    \(3x=0+15\)

                     =>    \(3x=15\)

                     =>    \(x=15:3\)

                     =>    \(x=5\)

\(\left(3x-5\right)^7=0\)

\(\Rightarrow3x-5=0\)

\(\Rightarrow3x=5\)

\(\Rightarrow x=\frac{5}{3}\)

Vậy xét là \(\frac{1}{2}+1\)nhé.

a,\(\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{1000}{999}\)

=3x4x5x...x1000/2x3x4x...x999

=1000/2=500

b, c tương tự câu a

)(1/2+1)x(1/3+1)x(1/4+1)x...x(1/999+1) 

b)(1/2-1)x(1/3-1)x(1/4-1)x...x(1/1000-1)

c)3/2² x 8/3² x 15/4² x .... x 99/10²

mình ko biết làm chép lại de thui

tự tính đi nhớ

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)

=> x+1 = 2003

=> x = 2003 - 1

=> x = 2002

thank you very much