Các câu hỏi tương tự
Giúp tui =((
\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
a, Tính : frac{left(13frac{1}{4}-2frac{5}{27}-10frac{5}{6}right).230frac{1}{25}+46frac{3}{4}}{left(1frac{3}{10}+frac{10}{3}right):left(12frac{1}{3}-14frac{2}{7}right)}b, Tính : Pfrac{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+100right)left(frac{1}{2}-frac{1}{3}-frac{1}{7}-frac{1}{9}right)left(63.1,2-21.3,6right)}{1-2+3-4+...+99-100}
Đọc tiếp
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 : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Rút gọn B= \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+5-6+...+99-100}\)
Rút gọn A=\(\frac{\left(1+2+3+......+99+100\right).\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right).\left(63.1,2-21.3,6+1\right)}{1-2+3-4+5-6+.........+99-100}\)=...
Tính:a.A frac{left(13frac{1}{4}-2frac{5}{27}-10frac{5}{6}right).230frac{1}{25}+46frac{3}{4}}{left(1frac{3}{10}+frac{10}{3}right):left(12frac{1}{3}-14frac{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+100right).left(frac{1}{2}-frac{1}{3}-frac{1}{7}-frac{1}{9}right).left(63.1,2-21.3,6right)}{1-2+3-4+...+99-100}
Đọc tiếp
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 = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Tính nhanh
\(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
6 tháng 3 2020 lúc 14:49
1.Chứng tỏ rằng:A75.(42004+42003+...+42+4+1)+25 chia hết cho 1002.tính nhanh:Afrac{left(1+2+3+...+99+100right)left(frac{1}{2}-frac{1}{3}-frac{1}{7}-frac{1}{9}right)left(63.1,2-21.3,6right)}{1-2+3-4+...+99-100}Bfrac{left(frac{1}{14}-frac{sqrt{2}}{7}+frac{sqrt[3]{2}}{35}right).left(-frac{4}{15}right)}{left(frac{1}{10}+frac{sqrt[3]{2}}{25}-frac{sqrt{2}}{5}right).frac{5}{7}}3.a)tính giá trị của biểu thức A3x2-2x+1 với |x|frac{1}{2}b)Tìm x nguyên để sqrt{x+1}chia hết cho sqrt{x-3}
Đọc tiếp
1.Chứng tỏ rằng:
A=75.(42004+42003+...+42+4+1)+25 chia hết cho 100
2.tính nhanh:
\(A=\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
\(B=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{\sqrt[3]{2}}{35}\right).\left(-\frac{4}{15}\right)}{\left(\frac{1}{10}+\frac{\sqrt[3]{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
3.a)tính giá trị của biểu thức A=3x2-2x+1 với |x|=\(\frac{1}{2}\)
b)Tìm x nguyên để \(\sqrt{x+1}\)chia hết cho \(\sqrt{x-3}\)
Rút gọn A=\(\frac{\left(1+2+3+.......+99+100\right).\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right).\left(63.1,2-21.36+1\right)}{1-2+3-4+5-6+....+99-100}\)là .....
tính
\(\frac{\left(1+2+3+....+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63\times1,2-21\times3,6\right)}{1-2+3-4+....+99-100}\)