Cho S=\(\dfrac{1}{5^2}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{2017}{5^{2017}}+\dfrac{2018}{5^{2018}}\).Chứng minh S<\(\dfrac{1}{3}\)
\(Tính: B = \dfrac{2 - \dfrac{2}{19} + \dfrac{2}{43} - \dfrac{2}{2017}}{3 - \dfrac{3}{19} + \dfrac{3}{43} - \dfrac{3}{2017}} :\dfrac{4 - \dfrac{4}{29} + \dfrac{4}{41} - \dfrac{4}{2018}}{5 - \dfrac{5}{29} + \dfrac{5}{41} - \dfrac{5}{2018}} \)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(B=\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(B=\dfrac{2}{3}:\dfrac{4}{5}\) ( Do \(\left\{{}\begin{matrix}1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\ne0\\1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\ne0\end{matrix}\right.\))
\(B=\dfrac{2}{3}\cdot\dfrac{5}{4}=\dfrac{2\cdot5}{3\cdot4}=\dfrac{5}{6}\)
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(\Rightarrow\)\(B=\dfrac{2-\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(\Rightarrow B=\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{10}{12}=\dfrac{5}{6}\)
tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
Tính
a, \(5\dfrac{4}{13}.14\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}\)
b, \(\dfrac{2^3}{3^3}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.2017^0\)
c, \(3:\left(\dfrac{-3}{2}\right)^2+\dfrac{1}{9}.\sqrt{36}\)
a, \(5\dfrac{4}{13}.15\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}\)
\(=\left(15\dfrac{3}{41}-2\dfrac{3}{41}\right).\dfrac{69}{13}=\dfrac{13.69}{13}=69\)
b, \(\dfrac{2^3}{3^3}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.2017^0\)
\(=\dfrac{8}{27}:\dfrac{16}{27}+\dfrac{2017}{2018}-\dfrac{1}{2}.1=\dfrac{1}{2}+\dfrac{2017}{2018}-\dfrac{1}{2}=\dfrac{2017}{2018}\)
c, \(3:\left(-\dfrac{3}{2}\right)^2+\dfrac{1}{9}.\sqrt{36}=3:\dfrac{9}{4}+\dfrac{1}{9}.6=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)
a,
\(5\dfrac{4}{13}.14\dfrac{3}{41}-5\dfrac{4}{13}.2\dfrac{3}{41}=5\dfrac{4}{13}.\left(14\dfrac{3}{41}-2\dfrac{3}{41}\right)\)
=\(5\dfrac{4}{13}.13\)
=\(\dfrac{69}{13}.13\)
=69
Chứng mình rằng A < 1 với A= \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}...+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2018^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)
=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=\dfrac{1}{2}-\dfrac{1}{2019}< 1\)
Vậy A < 1.
Cho A= 1 + \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{4034}\); B = 1 + \(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4033}\)
So sánh \(\dfrac{A}{B}\)với 1\(\dfrac{2017}{2018}\)
1.Tính
\(\left(1-\dfrac{1^2}{100}\right)\left(1-\dfrac{2^2}{100}\right)\left(1-\dfrac{3^2}{100}\right)...\left(1-\dfrac{2018^2}{100}\right)\)
2.S=\(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\)
Chứng minh rằng S <\(\dfrac{1}{2}\)
Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
a)Cho A= \(\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2021}{2015}\)
Chứng minh A>6
b)Cho C=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+....+\dfrac{1}{3^{2010}}\)
Chứng minh rằng C<1
Cho D=\(\dfrac{1}{1^2.2^3}+\dfrac{5}{2^2.3^3}+\dfrac{7}{3^2.4^2}+.....+\dfrac{4019}{2009^2.2010^2}\)
Chứng minh rằng D<1
mấy bạn giúp mình nha. Mình cần gấp lắm TT^TT
mấy bạn ơi câu b) là chứng minh C<\(\dfrac{1}{2}\)nha
BT1: Cho A = \(\dfrac{1}{2017}+\dfrac{2}{2017^2}+\dfrac{3}{2017^3}+...+\dfrac{2017}{2017^{2017}}+\dfrac{2018}{2017^{2018}}\)
Chứng minh rằng : A < \(\dfrac{2017}{2016^2}\)
Tính:
a) S=1.2+2.3+3.4+...+99.100
b) B=\(\dfrac{49^{24}.125^{17}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
c) C=\(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\right).3^5+\left(\dfrac{1}{3^5}+\dfrac{1}{3^6}+\dfrac{1}{3^7}+\dfrac{1}{3^8}\right).3^9+...+\left(\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right).3^{101}\)
d) D= \(3-3^2+3^3-3^4+...+3^{2017}-3^{2018}\)
a) S = 1.2 + 2.3 + 3.4 + ... + 99.100
S có thể được viết lại thành:
S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)
= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98
= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)
Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:
S = n(n+1)(2n+1)/6
Với n = 99, ta có:
S = 99.100.199/6 = 331650
Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:
S = n(n+1)/2
Với n = 98, ta có:
S = 98.99/2 = 4851
Do đó, S = 331650 - 4851 = 326799
b) B = 4924.12517.28−530.749.45529.162.748
B có thể được viết lại thành:
B = (4924.12517.28) / (530.749.45529.162.748)
B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)
B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529
B = 108 / 45529
c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101
C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101
C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)
C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)
C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)
d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018
D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^