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\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{101}\right)\)
\(=1.\frac{100}{101}\)
\(=\frac{100}{101}\)
\(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\)
\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
đặt biểu thức là A
5A=(1/1x6+1/6x11+...+1/96x101)x5=5/1x6+5/6x11+...+5/96x101
5A=6-1/1x6+11-6/6x11+...+101-96/96x101
5A=6/1x6-1/1x6+11/6x11-6/6x11+...+101/96x101-96/96x101
5A=1-1/6+1/6-1/11+...+1/96-1/101(sau khi rút gọn các phân số)
5A=1-1/101(còn lại sau khi trừ)
5A=100/101
A=100/101:5=20/101
Gọi A = 1/1.6 + 1/6.11 +...+ 1/(5n+1)(5n+6)
5A = 5/1.6 + 5/6.11 + ... + 5/(5n+1)(5n+6)
=1 - 1/6 + 1/6 - 1/11 + ... + 1/5n+1 - 1/5n+6
=1 - 1/5n+6 =5n+6/5n+6 - 1/5n+6=5n+5 /5n+6
1/1x2x3+1/2x3x4+...1/118x19x20<1/4 <--- cái này đề sai ở 1/118x19x20 phải là 1/18x19x20
a: Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
=>\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)
=>\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}\)
=>\(A=1-\dfrac{1}{32}=\dfrac{31}{32}\)
b: Đặt \(B=\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+...+\dfrac{1}{96\cdot101}\)
=>\(B=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{96\cdot101}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{5}\cdot\dfrac{100}{101}=\dfrac{20}{101}\)