Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\) . Chứng tỏ rằng \(S< \dfrac{1}{16}\)
Cho \(S=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{99}{5^{100}}\). Chứng tỏ rằng S<\(\dfrac{1}{16}\)
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
Chứng tỏ rằng :\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
Ta có :
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{2}{5}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+.................+\dfrac{99}{100}}\)
\(=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+.............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+............+1-\dfrac{1}{100}}\)
\(=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+...........+\dfrac{2}{100}\right)}{\left(1+1+.........+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+........+\dfrac{1}{100}\right)}\)
\(=\dfrac{2.\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..........+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.........+\dfrac{1}{100}\right)}\)
\(=2\)
Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+..........+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}=2\rightarrowđpcm\)
Chứng tỏ rằng: \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)=2
\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
Chứng tỏ rằng:
\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
A= \(\dfrac{1}{3}-\dfrac{2}{3^2}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
3A= 1 - \(\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+.....+\dfrac{99}{3^{98}}\) - \(\dfrac{100}{3^{99}}\)
A + 3A = 1- \(\dfrac{1}{3}+\dfrac{1}{3^2}\) - \(\dfrac{1}{3^3}+....+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}\)
=> 4A < 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}\) \(\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)
Đặt : B = 1 - \(\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+....+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)
3B = 3 - 1 + \(\dfrac{1}{3}\) - \(\dfrac{1}{3^2}+.....+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}\)
B + 3B = 3 - \(\dfrac{1}{3^{99}}\)
4B = 3 - \(\dfrac{1}{3^{99}}\) < 3 => B < \(\dfrac{3}{4}\)
=> 4A < \(\dfrac{3}{4}\) => A < \(\dfrac{3}{16}\) ĐPCM
Cho B=\(\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+...+\dfrac{2014}{5^{2015}}\). Chứng tỏ rằng B<\(\dfrac{1}{16}\)
a)chứng minh rằng :\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)........+\(\dfrac{1}{100^2}< \dfrac{1}{2}\)
b)tính nhanh tổng S với S= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+......+\dfrac{1}{61.63}\)
các cao nhân gải giúp với ạ !!! iem đang cần gấp
Câu 1 : Thực hiện phép tính 1 cách hợp lý :
a) \(\dfrac{-12}{7}.\dfrac{4}{35}+\dfrac{12}{7}.\dfrac{\left(-31\right)}{35}-\dfrac{2}{7}\)
b) \(1+2-3-4+5+5-7-8+...+97+98-99-100\)
c) \(A=157.\left(-37\right)-\left(41.53-37.157\right)+51.53\)
d) \(B=\left(\dfrac{1}{11}+\dfrac{1}{21}+\dfrac{1}{31}+\dfrac{1}{41}+\dfrac{1}{51}\right)\left(\dfrac{-41}{123}+\dfrac{31}{-186}-\dfrac{-51}{102}\right)\)
Câu 2 :
a) 12 ( x - 5 ) = 7x - 5
b) Tìm x \(\in\) Z sao cho : ( 2x - 3 ) 2010 = ( 2x - 3 ) 2012
Câu 3 :
1) Cho biểu thức S = 1 + 3 + 32 + 33 +...+ 3202 + 3 203
a) chứng tỏ rằng tổng S chia hết cho 52 .
b) Tìm Chữ số tận cùng trong tổng S .
2 ) Cho biểu thức A= \(\dfrac{2n+1}{2n+5}\) . Chứng tỏ rằng với mọi số tự nhiên n thì A là phân số tối giản .
Câu 4 : So sánh tổng gồm 1006 số hạng :
\(S=\dfrac{1}{1.1.3}+\dfrac{1}{2.3.5}+\dfrac{1}{3.5.7}+...+\dfrac{1}{1006.2011.2013}\) với \(\dfrac{2}{3}\)
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)