Giúp tui =((
\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Tính Nhanh
\(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Ta có: \(63.1,2-21.3,6=0,9.7.10.1,2-21.3,6\)
\(=6,3.12-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.12-6,3.12\)
\(=0\)
\(\Rightarrow\frac{\left(1+2+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)0}{1-2+3-4+...+99-100}=0\)
Vậy \(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=0\)
Dãy tính trên có (63.1,2 - 21.3,6) =0 nên kết quả là 0
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 = 13 + 23 + 33 + ... + 1253
Rút gọn A=\(\frac{\left(1+2+3+......+99+100\right).\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right).\left(63.1,2-21.3,6+1\right)}{1-2+3-4+5-6+.........+99-100}\)=...
Rút gọn B= \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+5-6+...+99-100}\)
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 : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
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:
A= \(\frac{\left(1+2+3+....+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3.6\right)}{1-2+3-4+...+99-100}\)
\(A=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63\cdot1,2-21\cdot3,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(75,6-75,6\right)}{1-2+3-4+...+99-100}\)
\(=\frac{\left(1+2+3+...+99+100\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\cdot0}{1-2+3-4+...+99-100}\)
\(=0\)
Vũ Minh Tuấn giúp mình với
Tính nhanh
\(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
Ta thấy : \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
nên kết quả dãy trên bằng 0
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 \(\sqrt{x-3}\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
3a) |x| = 1/2
=> x = 1/2 hoặc x = -1/2
với x = 1/2:
A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)
\(A=\frac{3}{4}-1+1=\frac{3}{4}\)
với x = -1/2
A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)
\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)
2.
A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\
A=0