Tìm x biết:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
Tìm x biết:
\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
\(\Rightarrow\frac{x-1}{65}-1+\frac{x-3}{63}-1=\frac{x-5}{61}-1+\frac{x-7}{59}-1\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}=\frac{x-66}{61}+\frac{x-66}{59}\)
\(\Rightarrow\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}=0\)
\(\Rightarrow x-66=0\).Do\(\frac{x-66}{65}+\frac{x-66}{63}-\frac{x-66}{61}-\frac{x-66}{59}\ne0\)
\(\Rightarrow x=66\)
Tìm x biết:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
x=66 nhé bn mk với
học tốt nhé ! nhớ kb
tìm x biết
\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
\(\frac{128x-258}{4095}=\frac{120x-722}{3599}\)
\(\left(128x-258\right)3599=4095\left(120x-722\right)\)
\(=>x=66\)
Giải pt: \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
Tìm x, biết:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\frac{x+29}{31}-\frac{x +17}{33}=\frac{x+17}{43}-\frac{x+15}{45}\)
Tìm số hữu tỉ x, biết:
a, \(\frac{x-100}{24}+\frac{x-98}{26}+\frac{x-96}{28}=3\)3
b, \(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
c, \(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)
giúp với gấp lắm
giải các bất phương trình:
\(\frac{x+1}{65}+\frac{x+3}{63}< \frac{x+5}{61}+\frac{x+7}{59}\)
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4< =2x-1\)
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 \(x>-66\)
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 :')
Tìm x :
\(\frac{x+1}{65}\)+ \(\frac{x+3}{63}\)=\(\frac{x+5}{61}\)+\(\frac{x+7}{59}\)
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
<=>\(\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)
<=>\(\frac{x+1}{65}+1+\frac{x+3}{63}+1-\frac{x+5}{61}-1-\frac{x+7}{59}-1=0\)
<=>\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
<=>\(\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Do \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
=>x+66=0
<=>x=-66
x= 66
nhe sbn
chắc chắn
thôi bn ạ
mình tính nháp rùi
Tìm x:
\(\frac{x-1}{65}+\frac{x-3}{66}=\frac{x-5}{61}+\frac{x-7}{59}\)
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