\(A=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+...+\frac{1}{3^{20}}\)
Những câu hỏi liên quan
1 CMR:
B=\(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+.....+\frac{3n+1}{3^n}< \frac{11}{4}\)(n thuộc N*;n>3)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
C=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^{20}-1}{3^{20}}>19\frac{1}{2}\)
Có : \(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow2A< 1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
Có: \(6A< 3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(6A-2A< 3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow4A< 3\Rightarrow A< \frac{3}{4}\)(đpcm)
Đúng 0
Bình luận (0)
Tính
a. S= 3 + \(\frac{3}{2}+\frac{3}{2^2}+......+\frac{3}{2^9}\)
b. A= \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}\)
c.M=\(\frac{20}{112}+\frac{20}{280}+\frac{20}{520}+\frac{20}{832}\)
Tính A = \(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+...+\frac{2+3+4+...+20}{1+2+3+4+...+20}\)
A=\(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+...+\frac{2+3+...+20}{1+2+3+...+20}\)
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.( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+...+ \(\frac{1}{17}\)+ \(\frac{1}{18}\)+ \(\frac{1}{19}\)+ \(\frac{1}{20}\))
=> \(\frac{Tử}{Mâu}\)= \(\frac{1}{20}\)
Đúng 0
Bình luận (0)
Phùng Quang Thịnh biến đổi sai 1 chỗ kìa
-19 = \(\frac{20}{20}-20\)chứ mà bạn
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
bài 1: tính nhanh
a, \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}\)-\(\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)
b,\(1\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{3}{20}+...+\frac{3}{2011.2012}\)
a,Ta có \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}-\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)
\(=\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{2.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)}-\frac{3.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}{6.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}\)
=\(\frac{1}{2}-\frac{3}{6}=\frac{1}{2}-\frac{1}{2}=0\)
Vậy giá trị biểu thức bằng 0
b, Mình không hiểu cho lắm ạ , nếu ko phiền xin xem lại đầu bài ạ
Đúng 0
Bình luận (0)
Bài 1:Tìm xa) b) c) d) e)Bài 2:Tính giá trịA B C DE FBài 3:TínhA a - b + c biết B a - b - c với
Đọc tiếp
Bài 1:Tìm x
a) b)
c)
d)
e)
Bài 2:Tính giá trị
A= B=
C=
D=
E= F=
Bài 3:Tính
A = a - b + c biết B = a - b - c với
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
Đúng 0
Bình luận (0)
left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
Đọc tiếp
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
1. CMR:
C = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}< \frac{1}{2}\)\(\frac{1}{2}\)
2. So sánh
a) 9920 và 99910
b) 920 và 2713
Tính giá trị của biểu thức \(A=\left(\frac{1}{125}-\frac{1}{1^3}\right).\left(\frac{1}{125}-\frac{1}{2^3}\right).\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{19^3}\right).\left(\frac{1}{125}-\frac{1}{20^3}\right)\)