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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
\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
\(\Rightarrow\frac{x-1}{61}-1\frac{x-3}{63}+-1=\frac{x-5}{61}-1+\frac{x-7}{59}-1\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Rightarrow\left(x-66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Rightarrow x=66\)
\(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\dfrac{x-1}{65}-1+\dfrac{x-3}{63}-1=\dfrac{x-5}{61}-1+\dfrac{x-7}{59}-1\)
\(\Leftrightarrow\dfrac{x-66}{65}+\dfrac{x-66}{63}=\dfrac{x-66}{61}+\dfrac{x-66}{59}\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
\(\Leftrightarrow x-66=0\)
\(\Leftrightarrow x=66\)
Vậy x=66.
\(\Rightarrow\frac{x-1}{65}-1+\frac{x-3}{63}-1=\frac{x-5}{61}-1+\frac{x-7}{59}-1\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}=0\)
\(\Rightarrow x-66=0\).Do\(\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}\ne0\)
\(\Rightarrow x=66\)
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{\left(x+1\right)}{65}-1+\frac{\left(x+3\right)}{63}-1=\frac{\left(x+5\right)}{61}-1+\frac{\left(x+7\right)}{59}\)
\(\Leftrightarrow\left(x-66\right).\left(\frac{1}{65}+\frac{1}{63}\right)=\left(x-66\right).\left(\frac{1}{61}+\frac{1}{59}\right)\)
\(\Leftrightarrow\left(x-66\right).\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Rightarrow x=66\)
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
=> \(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)=\left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)
=> \(\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
=> \(\left(x+66\right).\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
=> x + 66 = 0
=> x = 0 - 66
=> x = -66
(x-100)/24 + (x-98)/26 + (x-96)/28 = 3
<=> (x - 100)/24 -1 + (x-98)/26-1 (x-96)/28 -1 = 0
<=>(x-124)/24 + (x-124)/26 + (x - 124)/28 =0
<=>(x - 124) (1/24+1/26+1/28) = 0
vì 1/24+1/26+1/28 khác 0
=> x - 124 = 0
=> x = 124
2) (x-1)/65 + (x-3)/63 = (x-5)/61 + (x-7)/59
tương tự:
(x-1)/65 -1 +(x -3)/63 -1 = (x-5)/61-1 + (x-7)/59 -1
rút gọn được:
(x - 66).(1/65 + 1/63) = (x -66).(1/61 + 1/59)
(x - 66).(1/65 + 1/63 - 1/61 -1/59) = 0
=> x = 66 (lý luận tương tự câu trên)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)
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 \(x\in\left\{-66\right\}\)