\(\frac{7256.4375-725}{3650+4375.7255}=\frac{\left(7255+1\right).4375-725}{3650+4375.7255}\)
\(=\frac{7255.4375+4375-725}{3650+4375.7255}\)
\(=\frac{7255.4375+3650}{3650+4375.7255}\)
\(=1\)
\(\frac{7256.4375-725}{3650+4375.7255}=\frac{\left(7255+1\right).4375-725}{3650+4375.7255}\)
\(=\frac{7255.4375+4375-725}{3650+4375.7255}\)
\(=\frac{7255.4375+3650}{3650+4375.7255}\)
\(=1\)
a)\(\left(5.4^{11}-3.16^5\right):4^{10}\)
b)\(\dfrac{7256.4375-725}{3650+4375.7255}\)
c)100+98+96+...+4+2-97-95-...-3-1
giúp mình nhé
1.Tính nhanh
A,7256×4375-725/3650+4375×7255
Tìm chữ số tận cùng của các tổng sau:
a) A= 225+389+8413+5617
b) B= 762+1014+7258+57612
c) C= 3910+7214+3318+6422
Ai giúp mình sẽ tick cho người đó
Tìm chữ số tận cùng của các tổng sau:
a) A= 225+389+8413+5617
b) B= 762+1014+7258+57612
c) C= 3910+7214+3318+6422
Ai giúp mình sẽ tick cho người đó
Rút gọn
1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
2. \(\left(\frac{1+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}}{2+\frac{2}{5}+\frac{2}{7}+\frac{2}{17}}:\frac{4-\frac{4}{7}+\frac{4}{9}-\frac{4}{13}}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{13}}\right):\frac{838383}{808080}\)
Rút gọn
1.\(\left(\frac{2}{45}-\frac{4}{13}-\frac{1}{3}\right):\left(\frac{3}{13}-\frac{4}{15}+\frac{2}{13}\right)\)
2.\(\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{15}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)2\frac{2}{17}}\)
3.\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
Thực hiên phép tính:
\(\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{37}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
Chứng minh :
a) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{4^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
b)\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}< \frac{7}{12}\)
c) Cho \(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)
Chứng minh \(1< S< 2\)
Rút gọn
1.\(\frac{7}{19}.\frac{8}{11}+\frac{7}{19}.\frac{3}{11}+\frac{12}{19}\)
2.\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)