C1:
Ta có
x+10%x-10%x=297
=>x=297
C2:
S=Đề bài...
=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)
=\(\frac{1}{1}-\frac{1}{101}\)
=\(\frac{100}{101}\)
#hoctot
Câu 2:
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\)
\(\Rightarrow2S=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow2S=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.100}\)
\(\Rightarrow2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}+\frac{1}{101}\)
\(\Rightarrow2S=1-\frac{1}{101}\)
\(\Rightarrow2S=\frac{100}{101}\)
\(\Rightarrow S=\frac{100}{101}:2=\frac{100}{101}.\frac{1}{2}=\frac{50}{101}\)
\(S=\frac{1}{1.3}+....+\frac{1}{99.100}\)
\(S=1-\frac{1}{100}\\ S=\frac{99}{100}\)