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mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)
\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)
\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)
Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=0-66=-66\)
Auto làm nốt câu b
a, Cộng cả 2 vế với 2
Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)
\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)
=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)
Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)
=> \(x=-65\)
b , Lm tương tự như Câu a
Chúc bn hok tốt
\(a/\frac{7}{9}-\frac{x}{3}=\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{9}-\frac{1}{9}\)
\(\Rightarrow\frac{x}{3}=\frac{2}{3}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(b/\frac{1}{x}-\frac{-2}{15}=\frac{7}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{7}{15}+\frac{-2}{15}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{3}\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
\(c/\frac{-11}{14}-\frac{-4}{x}=\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-11}{14}-\frac{-3}{14}\)
\(\Rightarrow\frac{-4}{x}=\frac{-4}{7}\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
\(d/\frac{x}{21}-\frac{2}{3}=\frac{5}{21}\)
\(\Rightarrow\frac{x}{21}=\frac{5}{21}+\frac{2}{3}\)
\(\Rightarrow\frac{x}{21}=\frac{19}{21}\)
\(\Rightarrow x=19\)
Vậy \(x=19\)
#Mạt Mạt#
a) \(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
\(\frac{3}{4}:x=\frac{3}{8}\)
\(x=2\)
vậy x=2
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
\(x=2001\)
vậy x=2001
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