Cho :
A = \(\left(\frac{1}{2}_{ }+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+.....+\frac{1}{2016}+\frac{1}{2017}\right)\)
B = \(\left(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Tính \(\frac{B}{A}\) ?
[Các bạn giúp mình với !!!]
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
tính
A=\(\left(\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}\right)\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)
\(\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}+1\right)\left(\frac{2105}{2016}+\frac{2016}{2017}+\frac{7}{22}\right)-\left(\frac{1}{2}+\frac{2015}{2016}+\frac{2016}{2017}\right)\left(\frac{2015}{2016}+\frac{2016}{2017}+\frac{7}{22}+1\right)\)
Tính B = \(\left(\frac{1}{1009}+\frac{1}{1010}+....+\frac{1}{2015}+\frac{1}{2016}\right):\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{2016}\right)\)
Tính giá trị các biểu thức sau:
a,A=\(\frac{2}{3}+\frac{5}{6}:5-\frac{1}{18}.\left(-3\right)^2\)
b,B=\(3.\left(5.\left(\left(5^2+2^3\right):11\right)-16\right)+2015\)
c,C=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
Tìm giá trị của biểu thức sau:
a)\(A=\frac{2}{3}+\frac{5}{6}:5-\frac{1}{18}.\left(-3\right)^2\)
b)\(B=3.\left\{5.\left[\left(5^2+2^3\right):11\right]-16\right\}+2015\)
c)\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2014.2016}\right)\)
https://dethihsg.com/de-thi-hoc-sinh-gioi-phong-gddt-hoang-hoa-2014-2015/
vào đây gợi ý nhé
k mik đi
@_@
Bài 1 : Tính :
a)\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\times230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right)\div\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) \(\frac{2^{12}\times3^5-4^6\times9^2}{\left(2^4\times3\right)^6+8^4\times3^5}-\frac{5^{10}\times7^3-25^5\times49^2}{\left(125\times7\right)^3+5^9\times14^3}\)
c)P=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+....+\frac{1}{2015}}\)
Tính:
a/\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
b/\(\frac{2\cdot2306}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+230}}\)c/\(\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+...+\frac{1}{44\cdot49}\right)\left(\frac{1-3-5-...-49}{89}\right)\)
\(b.\)ghi lại đề nha bn
\(=\frac{2.2306}{1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{230.231}{2}}}\)
\(=\frac{2.2306}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{230.231}}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{230.231}\right)}\)
\(=\frac{2.2306}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{230}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+2.\left(\frac{1}{2}-\frac{1}{231}\right)}\)
\(=\frac{2.2306}{1+1-\frac{2}{231}}\)
\(=\frac{2.2306}{2-\frac{2}{231}}\)
\(=\frac{2.2306}{2\left(1-\frac{1}{231}\right)}\)
\(=\frac{2306}{1-\frac{1}{231}}\)
mình nha bn thanks nhìu <3
a) \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+...+\frac{1}{2016}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{\frac{2017}{2}+...+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2}+...+\frac{1}{2016}+\frac{1}{2017}\right)}\)
\(=\frac{1}{2017}\)
