Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
y =2(\(\frac{37\cdot10101}{120\cdot10101}\)-\(\frac{37\cdot10101}{6\cdot10101}\)):\(\frac{1}{x}\)=-37
y=2(37/120-37/6):1/x=-37
y=2(37/120-740/120):1/x=-37
y=2.(-703/120):1/x=-37
y=--703/60:1/x=-37
1/x=-703/60:-37
1/x=19/60
x=60/19
A=\(\frac{63x101x37-37x101x63}{1+2+3+..+2014}\)(vì 636363=63x10101 và 373737=37x101x63)
A\(\frac{0}{1+2+3+4...+2014}\)
A=0
\(A=\frac{636363\times37-373737\times63}{1+2+3+....+2014}\)
\(A=\frac{63\times10101\times37-373737\times63}{1+2+3+......+2014}\)
\(A=\frac{373737\times63-373737\times63}{1+2+3+.....+2014}\)
\(A=\frac{0}{1+2+3+.....+2014}\)
\(A=0\)
HOK TOT
Mình nghĩ đề thế này mới tính hợp lí được
2 ) B = \(1\frac{6}{41}.\left(\frac{12+\frac{12}{19}-\frac{12}{37}-\frac{12}{53}}{3+\frac{3}{19}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2006}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2006}}\right).\frac{124242423}{237373735}\)
B = 47/41 . ( 12/3 : 4/5 ) . 123/235
B = 47/41 . ( 4 : 4/5 ) . 123/235
B = 47/41 . 5 . 123/235
B = \(\frac{47.5.123}{41.235}\)
B = 3
1 ) A = \(\frac{636363.37-373737.63}{1+2+3+...+2006}\)
A = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2006}\)
A = \(\frac{0}{1+2+3+...+2006}\)
A = 0
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
Ta có :
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{10^2}\right)\)
\(=\)\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.....\frac{10^2-1}{10^2}\)
\(=\)\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{9.11}{10.10}\)
\(=\)\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)
\(=\)\(\frac{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}\)
\(=\)\(\frac{11}{2.10}\)
\(=\)\(\frac{11}{20}\)
Chúc bạn học tốt ~
(1-1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/10^2)
=3/4.8/9.15/16...99/100
Từ đó tính kết quả là ok
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{49}{50}\)
\(=\frac{1}{50}\)
Chỗ nào không hiểu nhắn tin cho tớ nha!
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{49}{50}\)
\(=\frac{1}{50}\)