\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)

\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)

\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)

Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\) 

\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)

Vậy để \(\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)

\(\Leftrightarrow x+66=0\)

\(\Leftrightarrow x=-66\)

Vậy \(x\in\left\{-66\right\}\)

mỗi phân số cộng thêm 1 ta có

\(\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)

rồi đặt x+66 làm thừa số chung sau đó giải tiếp

x thuộc tập hợp rỗng

\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)

\(\Rightarrow\frac{x-1}{61}-1\frac{x-3}{63}+-1=\frac{x-5}{61}-1+\frac{x-7}{59}-1\)

\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)

\(\Rightarrow\left(x-66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)

\(\Rightarrow x=66\)

1+1+2+3+5+8.......+89

Ta se ket hop cho dau tien vao

1+1=2

2+2=4

4+3=7

7+5=12

12+8=20

Vay chung ta se co ket qua la 20

Lay 89 tru 20 bang 69 na

Vay : 1+1+2+3+5+8+69+89 = 178

Dung thi k cho minh con sai thi thoi nha

Minh hoc lop 3

Bài 2: 

Ta có: \(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)

\(\Leftrightarrow\left(\dfrac{x-1}{65}-1\right)+\left(\dfrac{x-3}{63}-1\right)=\left(\dfrac{x-5}{61}-1\right)+\left(\dfrac{x-7}{59}-1\right)\)

\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)

=>x-66=0

hay x=66