tính
\(B=\frac{6}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:1\frac{1}{2011}:\left(-1\frac{1}{2012}\right)\)
Thực hiện phép tính:
a) \(\left(\frac{1}{3}-1\right).\left(\frac{1}{6}-1\right).\left(\frac{1}{10}-1\right).\left(\frac{1}{15}-1\right)...\left(\frac{1}{1225}-1\right).\left(\frac{1}{1275}-1\right)\)
b) 1 - 2 + 3 - 4 + 5 - 6 + ... + 2011 - 2012
B
từ 1 đến 2012 có tất cả:
2012-1:1+1 = 2012 (số)
=>có: 2012:2 = 1006 (cặp)
Mà mỗi cặp bằng (-1)nên
tổng dãy số trên là: 1006 . (-1) = -1006
(1-2)+(2-3)+(3-4)+(5-6)+...+(2011-2012)
=-1+(-1)+(-1)+(-1)+...+(-1)
có tất cả các số -1 trên dãy số trên là
(2012-2);2+1=1006
vậy suy ra ; -1x1006=(-1006)
chac chan la dung
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Bài 3 : a) Tính
\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b) Tính :
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+\frac{1}{2011}}\)
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
GPT :
a, \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}..\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
b, \(\left(\frac{0,6+\frac{3}{7}-\frac{2}{11}}{1+\frac{5}{7}-\frac{5}{11}}+\frac{\frac{2}{3}-1,5+\frac{2}{9}}{\frac{5}{3}-3,75+\frac{5}{9}}\right)+93x=\left(\frac{3737}{4545}-\frac{954954}{975975}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)-7.\left(x-3\right)\)
Tính:
a.A = \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b. B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c. C = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Câu 1 Tính giá trị biểu thức:
a) A= \(2+5+8+11+...+2012\)
b) B=\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2011}\right).\left(1-\frac{1}{2012}\right)\)
Câu 2
a) Tìm x,y nguyên biết : 2x.(3y-2) + (3y-2) = -55
b) CMR:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
Câu 1 :
A = (2012+2) . [ ( 2012-2) : 3+1 ] : 2 = 2014 . 671 : 2 = 675697
B = \(\frac{1}{2}\). \(\frac{2}{3}\). \(\frac{3}{4}\)+...+ \(\frac{2010}{2011}\). \(\frac{2011}{2012}\)= \(\frac{1.2.3.....2010.2011}{2.3.4.....2011.2012}\)= \(\frac{1}{2012}\)
Câu 2 :
a) \(2x.\left(3y-2\right)+\left(3y-2\right)=-55\)
=> \(\left(3y-2\right).\left(2x+1\right)=-55\)
=> \(3y-2;2x+1\in\: UC\left(-55\right)\)
=> \(3y-2;2x+1=\left\{1;-1;5;-5;11;-11;55;-55\right\}\)
- Vậy ta có bảng
\(2x+1\) | 1 | -1 | 5 | -5 | 11 | -11 | 55 | -55 |
\(x\) | 0 | -1 | 2 | -3 | 5 | -6 | 27 | -28 |
\(3y-2\) | -55 | 55 | -11 | 11 | -5 | 5 | -1 | 1 |
\(3y\) | -53 | 57 | -9 | 13 | -3 | 7 | 1 | 3 |
\(y\) | \(\frac{-53}{3}\)(loại) | 19(chọn) | -3(chọn) | \(\frac{13}{3}\)(loại) | -1(chọn) | \(\frac{7}{3}\)(loại) | \(\frac{1}{3}\)(loại) | 1(chọn) |
\(\Leftrightarrow\)Những cặp (x;y) tìm được là :
(-1;19) ; (2;-3) ; (5;-1) ; (-28;1)
b) Ta đặt vế đó là A
Ta xét A : \(\frac{1}{4^2}\)< \(\frac{1}{2.4}\)
\(\frac{1}{6^2}\)< \(\frac{1}{4.6}\)
\(\frac{1}{8^2}\)< \(\frac{1}{6.8}\)
...
\(\frac{1}{\left(2n\right)^2}\)< \(\frac{1}{\left(2n-2\right).2n}\)
\(\Leftrightarrow\)A < \(\frac{1}{2.4}\)+ \(\frac{1}{4.6}\)+...+ \(\frac{1}{\left(2n-2\right).2n}\)
\(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{2}{2.4}\)+ \(\frac{2}{4.6}\)+...+ \(\frac{2}{\left(2n-2\right).2n}\))
\(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{6}\)+...+ \(\frac{1}{2n-2}\)- \(\frac{1}{2n}\))
\(\Leftrightarrow\)A < \(\frac{1}{2}\). ( \(\frac{1}{2}\)- \(\frac{1}{2n}\)) = \(\frac{1}{2}\). \(\frac{1}{2}\)- \(\frac{1}{2}\). \(\frac{1}{2n}\)
\(\Leftrightarrow\)A < \(\frac{1}{4}\)- \(\frac{1}{4n}\)< \(\frac{1}{4}\) ( Vì n \(\in\)N )
\(\Leftrightarrow\)A < \(\frac{1}{4}\)( đpcm ) .
Bạn Phùng Quang Thịnh làm đúng hết rồi
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
BÀi 1: Thực hiện phép tính ( tính nhanh nếu có thể)
a.\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{71}-\left(-\frac{2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
b.\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
c.\(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{1}{3}-1\frac{1}{15}\right)\)