Ta có: \(\dfrac{1991.1993-1}{1990+1991.1992}\)
\(=\dfrac{1991\left(1992+1\right)-1}{1991.1992+1990}\)
\(=\dfrac{1991.1992+1991.1-1}{1991.1992+1990}\)
\(=\dfrac{1991.1992+1991-1}{1991.1992+1990}\)
\(=\dfrac{1991.1992+1990}{1991.1992+1990}=1\)
Vậy giá trị của biểu thức là \(1\)
Chứng tỏ:
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1995}-\dfrac{1}{1996}=\dfrac{1}{996}+\dfrac{1}{997}+...+\dfrac{1}{1990}\)
Giúp Mình vs
So sánh:
A = \(\dfrac{10^{1990}+1}{10^{1991}+1}\)
B = \(\dfrac{10^{1991}+1}{10^{1992}+1}\)
so sánh:
199010 + 19909 và 199110
So Sánh A=101990+1 phần 101991+1 và B= 101991+1 phần 101992+ 1
GIÚP VỚI MAI MÌNH THI HỌC KÌ RÙI AI NHANH TICK NHA !
BT1: CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)
b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)
d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
BT2: Tính tổng
a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)
BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
CMR: 1 < S < 2
\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1x99}+\dfrac{1}{3x97}+\dfrac{1}{5x95}+...+\dfrac{1}{97x3}+\dfrac{1}{99x1}}\)
19910 + 19909 và 199110
Tính:
\(\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{3}-1\right):\left(\dfrac{1}{4}-1\right):\) ... : \(\left(\dfrac{1}{50}-1\right)\)
Chứng minh rằng:
\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}+\dfrac{1}{102}\right)=\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}\)
1) Rút gọn
A =\(\dfrac{\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+.......+\dfrac{18}{2}+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.......+\dfrac{1}{19}+\dfrac{1}{20}}\)
2) Tìm x
a/ \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x.\left(x+1\right)}=\dfrac{2016}{2017}\)
