Phân tích mẫu thức thành nhân tử ta có : 

1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+7)(x+8)=1/14

1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+7)-1/(x+8)=1/14

1/(x+1)-1/(x+8)=1/14

7/(x+1)(x+8)=1/14

Nhân chéo ta có x^2+9x+8=98

x^2+9x-90=0

(x+15)(x-6)=0

Suy ra x=-15 hoặc x=6

sửa đề đi bạn. Đọc không ra

So sánh \(A=\frac{2017^{2016}+1}{2017^{2017}+1}\)và\(B=\frac{2017^{2017}+1}{2017^{2018}+1}\)
ta có: \(\left(2017^{2016}+1\right)\left(2017^{2018}+1\right)=2017^{2016+2018}+2017^{2016}+2017^{2018}+1\)
=\(2017^{4034}+2017^{2017}\cdot\frac{1}{2017}+2017^{2017}\cdot2017+1=2017^{4034}+2017^{2017}\left(\frac{1}{2017}+2017\right)+1\)
         \(\left(2017^{2017}+1\right)\left(2017^{2017}+1\right)=2017^{4034}+2\cdot2017^{2017}+1\)
Vì \(2017+\frac{1}{2017}>2\)nên\(2017^{4034}+2017^{2017}\left(2017+\frac{1}{2017}\right)+1>2017^{4034}+2\cdot2017^{2017}+1\)
\(\Rightarrow\left(2017^{2016}+1\right)\left(2017^{2018}+1\right)>\left(2017^{2017}+1\right)\left(2017^{2017}+1\right)\)
\(\Rightarrow\frac{2017^{2016}+1}{2017^{2017}+1}>\frac{2017^{2017}+1}{2017^{2018}+1}\)
\(\Rightarrow A>B\)

Ta có: x²-y² = (8x+y) - (8y+x)

    => (x-y)(x+y) = 7x - 7y (Hằng đẳng thức nhé!)

         (x-y)(x+y) = 7(x-y)

   => x+y = 7 (cùng chia cả 2 vế cho x-y)

Ta có: x²+y² = (8x+y) + (8y+x)

          x²+y² = 9x + 9y

         x²+y² = 9(x+y)    (*)

Thay x+y = 7 vào biểu thức (*) ta được

         x²+y² = 63

Vậy x²+y² = 63