\(\text{Ta có: }A=x^{2005}-2006x^{2004}+2006x^{2003}-2006x^{2002}+...-2006x^2+2006x-1.\)\(=x^{2005}-\left(2005+1\right)x^{2004}+\left(2005+1\right)x^{2003}-\left(2005+1\right)x^{2002}+...-\left(2005+1\right)x^2+\left(2005+1\right)x-1\)  \(\text{Mà x=2005 nên: }A=x^{2005}-x^{2005}-x^{2004}+x^{2004}+x^{2003}-x^{2003}-x^{2002}+...-x^3-x^2+x^2+x-1\)

  \(=x-1=2005-1=2004\)

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x + 1) = 99/100
1- 1/2 +1/2-1/3+1/3-1/4+...+ 1/x - 1/ x+ 1 = 99/100
1 - 1/ x+1 = 99/ 100
=> (100 - 1)/ x+1 = 99 / 100
=> x+1 = 100 => x=99

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\)

mình sửa đề mới làm đc cái chỗ 1/2 phải là 1/1.2 ( đúng ko . xem lại )

A = 1/1.2 + 1/2.3 +....+ 1/x.(x+1)=99/100

A=1/1 - 1/2 + 1/2 - 1/3 +...+ 1/x - 1/x+1 =99/100

A = 1 - 1/x+1 = 99/100

A=x+1 - 1/x+1 = 99/100

A=x/x+1 = 99/100

=> x=99

\(\frac{2006x2005-1}{2004x2006+2005}\)

=\(\frac{2005-1}{2004+2005}\)

=\(\frac{2004}{4009}\)

mik ko hiểu nguyễn tạo nguyên ơi

OKKKKK

\(x-2xy+y=0\)

\(\Rightarrow x-\left(2xy-y\right)=0\)

\(\Rightarrow x-y\left(2x-1\right)=0\)

\(\Rightarrow2x-2y\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right)-2y\left(2x-1\right)=-1\)

\(\Rightarrow\left(2x-1\right)\left(1-2y\right)=-1\)

\(\Rightarrow\left(2x-1;1-2y\right)=\left(-1;1\right);\left(1;-1\right)\)

\(\Rightarrow\left(x;y\right)=\left(0;0\right);\left(1;1\right)\)