Cộng cả tử và mẫu với 1=> Kết quả là -66

Bạn cộng cả tử và mẫu vs 1 là xong 

-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 = -66 nha ban

k tui nha

thanks

x=-66 nha.

x=66 nhé bn mk với

             học tốt nhé ! nhớ kb

Ta có : 

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

\(\Leftrightarrow\)\(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)< \left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+5}{61}-\frac{x+7}{59}< 0\)

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

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

\(\Rightarrow\)\(x+66>0\)

\(\Rightarrow\)\(x>-66\)

Vậy \(x>-66\)

mình nhầm câu 2 phải là <=0

Ta có : 

\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4\le0\)

\(\Leftrightarrow\)\(\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)+4\le0\)

\(\Leftrightarrow\)\(\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}+4\le0\)

\(\Leftrightarrow\)\(\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)+4\le0\)

Hết biết giải, mk mới lớp 7 :')

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

\(\frac{128x-258}{4095}=\frac{120x-722}{3599}\)

\(\left(128x-258\right)3599=4095\left(120x-722\right)\)

\(=>x=66\)

Bài 15:

Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)

\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)

\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)

Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)

nên x+2010=0

hay x=-2010

Vậy: x=-2010

Bài 17:

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

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

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

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

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

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

Vì \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)

nên x+66=0

hay x=-66

Vậy: x=-66