giải pt sau
\(\frac{x+1}{65}\) +\(\frac{x+3}{65}\) =\(\frac{x+5}{16}\) \(\frac{x+7}{59}\)
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RB
4
1 tháng 4 2020
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
TL
3
3 tháng 7 2017
Ta có : \(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}=\frac{x-7}{59}\)
\(\Leftrightarrow\left(\frac{x-1}{65}-1\right)+\left(\frac{x-3}{63}-1\right)=\left(\frac{x-5}{61}-1\right)+\left(\frac{x-7}{59}-1\right)\)
\(\Leftrightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Leftrightarrow\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}=0\)
\(\Leftrightarrow\left(x-66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Mà ; \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)\ne0\)
Nên x - 66 = 0
=> x = 66
NN
1
TN
13 tháng 7 2016
\(\Rightarrow\frac{x-1}{65}-1+\frac{x-3}{63}-1=\frac{x-5}{61}-1+\frac{x-7}{59}-1\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}=0\)
\(\Rightarrow x-66=0\).Do\(\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}\ne0\)
\(\Rightarrow x=66\)
SG
2
LM
7 tháng 12 2017
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{\left(x+1\right)}{65}-1+\frac{\left(x+3\right)}{63}-1=\frac{\left(x+5\right)}{61}-1+\frac{\left(x+7\right)}{59}\)
\(\Leftrightarrow\left(x-66\right).\left(\frac{1}{65}+\frac{1}{63}\right)=\left(x-66\right).\left(\frac{1}{61}+\frac{1}{59}\right)\)
\(\Leftrightarrow\left(x-66\right).\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Rightarrow x=66\)
KK
5 tháng 1 2019
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
=> \(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)=\left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)
=> \(\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
=> \(\left(x+66\right).\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
=> x + 66 = 0
=> x = 0 - 66
=> x = -66
NT
7
13 tháng 2 2018
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 \(x>-66\)
TT
1
31 tháng 7 2015
\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
\(\frac{128x-258}{4095}=\frac{120x-722}{3599}\)
\(\left(128x-258\right)3599=4095\left(120x-722\right)\)
\(=>x=66\)
sê đài
Sửa đề:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\\\Leftrightarrow \frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\\ \Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\\\Leftrightarrow \left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\\\Leftrightarrow x+66=0\left(Vi\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)\\ \Leftrightarrow x=-66\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-66\right\}\)