Tính nhanh A = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+19}\)
\(\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{18}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{19}}\)
đây là bài yêu cầu tính nhanh nha
ta có
tử số \(\frac{1}{19}+\frac{2}{18}+..+\frac{18}{2}+\frac{18}{1}=\frac{1}{19}+1+\frac{2}{18}+1+..+\frac{18}{2}+1\)
\(\frac{20}{19}+\frac{20}{18}+..+\frac{20}{2}=20\left(\frac{1}{19}+\frac{1}{18}+..+\frac{1}{2}\right)\)
Do đó ta có phân số trên bằng 20
Tính
\(\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+......+\frac{18}{2}+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{19}+\frac{1}{20}}\)
Các bạn giúp mình nhanh lên mình đang cần gấp
Tính A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}}{\frac{19}{1}+\frac{18}{2}+\frac{17}{3}+...+\frac{3}{17}+\frac{2}{18}+\frac{1}{19}}\)
* Cách làm : Tử giữ nguyên,còn mẫu ta biến đổi như sau:
Mẫu : ( \(\frac{19}{1}\)+ 1 ) + ( \(\frac{18}{2}\)+ 1 ) + ( \(\frac{17}{3}\)+ 1 ) +...+ ( \(\frac{3}{17}\)+ 1 ) + ( \(\frac{2}{18}\)+ 1 ) + ( \(\frac{1}{19}\)+ 1 ) - 19 ( vì ta cộng với 19 số 1 nên phải trừ 19 )
= \(\frac{20}{1}\)+ \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)- 19
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ ( \(\frac{20}{1}\)- 19)
= \(\frac{20}{2}\)+ \(\frac{20}{3}\)+ ...+ \(\frac{20}{17}\)+ \(\frac{20}{18}\)+ \(\frac{20}{19}\)+ \(\frac{20}{20}\)
= 20.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
Tính nhanh các biểu thức sau:
a) A = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{19}\)
b) B = \(\frac{2}{3}+\frac{2}{6}+\frac{2}{9}+...+\frac{2}{90}\)
c) C = \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\)
Tính:
a) \(A=\frac{(1+17)(1+\frac{17}{2})(1+\frac{17}{3})...(1+\frac{17}{19})}{(1+19)(1+\frac{19}{2})(1+\frac{19}{3})...(1+\frac{19}{17})}\)
b) \(B=\frac{1}{-2}.\frac{1}{3}+\frac{1}{-3}.\frac{1}{4}+...+\frac{1}{-5}.\frac{1}{10}\)
c) \(C=(1-\frac{1}{1.2})+(1-\frac{1}{2.3})+...+(1-\frac{1}{2015.2016})\)
d) \(D=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+...+\frac{1}{10}}\)
Bài 1: Tính
\(\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+....+\frac{18}{2}+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{19}+\frac{1}{20}}\)
Tính
A=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}{\frac{19}{1}+\frac{18}{2}+\frac{17}{3}+...+\frac{1}{19}}\)
Tính :
P = \(\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+....+\frac{18}{2}+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}}\)
Ta có phần tử \(=\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{18}{2}+\frac{19}{1}\)
\(=\left(\frac{1}{19}+1\right)+\left(\frac{2}{18}+1\right)+...+\left(\frac{18}{2}+1\right)+\left(\frac{19}{1}+1\right)-19\)
\(=\frac{20}{19}+\frac{20}{18}+...+\frac{20}{2}+\frac{20}{1}+\frac{20}{20}-20\)
\(=20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}\right)\left(1\right)\)
Thay (1) vào P ta được :
\(P=\frac{20.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}}\)
\(=20\)
thực hiện phép tính (tính nhanh nếu có thể) :
a)\(1\frac{3}{19}+\frac{8}{21}-\frac{3}{19}+0.5+\frac{13}{21}\)
b)\(\left(-\frac{3}{4}+\frac{2}{7}\right):\frac{3}{7}+\left(\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
c)\(\left(\frac{1}{6}\right)^7\times6\)
d)\(\left(\frac{-1}{2}\right)^3+\frac{1}{2}:3\)
a) \(1\frac{3}{19}+\frac{8}{21}-\frac{3}{19}+0.5+\frac{13}{21}\)
\(=\left(1\frac{3}{19}-\frac{3}{19}\right)+\left(\frac{8}{21}+\frac{13}{21}\right)+0.5\)
\(=1+1+0.5=2.5\)
b) \(\left(-\frac{3}{4}+\frac{2}{7}\right):\frac{3}{7}+\left(\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=0:\frac{3}{7}=0\)