Bài 1: Tìm x
a) \(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
b) Bài 2: Chứng tỏ
\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
Mình cần gấp lắm mấy ban ưi ~~~
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
\(\frac{1}{4}+\frac{3}{4}:x=\frac{5}{8}\)
\(\frac{3}{4}:x=\frac{5}{8}-\frac{1}{4}\)
\(\frac{3}{4}:x=\frac{5}{8}-\frac{2}{8}\)
\(\frac{3}{4}:x=\frac{3}{8}\)
\(x=\frac{3}{4}:\frac{3}{8}\)
\(x=\frac{3}{4}.\frac{8}{3}\)
\(x=\frac{8}{4}\)
\(x=\frac{1}{2}=2\)
