Ôn tập toán 6
Các câu hỏi tương tự
Chứng tỏ rằng : \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
Chứng tỏ
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)
Tính:
left(dfrac{1}{2}-1right):left(dfrac{1}{3}-1right):left(dfrac{1}{4}-1right): ... : left(dfrac{1}{50}-1right)
Chứng minh rằng:
left(1+dfrac{1}{3}+dfrac{1}{5}+...+dfrac{1}{50}right)-left(dfrac{1}{2}+dfrac{1}{4}+dfrac{1}{6}+...+dfrac{1}{100}+dfrac{1}{102}right)dfrac{1}{52}+dfrac{1}{53}+...+dfrac{1}{100}+dfrac{1}{101}+dfrac{1}{102}
Đọc tiếp
Tính:
\(\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{3}-1\right):\left(\dfrac{1}{4}-1\right):\) ... : \(\left(\dfrac{1}{50}-1\right)\)
Chứng minh rằng:
\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}+\dfrac{1}{102}\right)=\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}\)
BT1: CMR:
a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{n^2} 1
b) dfrac{1}{4}+dfrac{1}{16}+dfrac{1}{36}+dfrac{1}{64}+dfrac{1}{100}+dfrac{1}{144}+dfrac{1}{196} dfrac{1}{2}
c) dfrac{1}{3}+dfrac{1}{30}+dfrac{1}{32}+dfrac{1}{35}+dfrac{1}{45}+dfrac{1}{47}+dfrac{1}{50} dfrac{1}{2}
d) dfrac{1}{2}-dfrac{1}{4}+dfrac{1}{8}-dfrac{1}{16}+dfrac{1}{32}-dfrac{1}{64} dfrac{1}{3}
e) dfrac{1}{3} dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+...+dfrac{99}{3^{99}}-dfrac{100}{3^{100}} dfrac{3}{16}
f) d...
Đọc tiếp
BT1: CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)
b) \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
c) \(\dfrac{1}{3}+\dfrac{1}{30}+\dfrac{1}{32}+\dfrac{1}{35}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{50}< \dfrac{1}{2}\)
d) \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
e) \(\dfrac{1}{3}< \dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)
f) \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
BT2: Tính tổng
a) A=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
b) E=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{200}\left(1+2+3+...+200\right)\)
BT3: Cho S=\(\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}+\dfrac{3}{13}+\dfrac{3}{14}\)
CMR: 1 < S < 2
Thu gọn các tổng sau:
a. A=8.5100.(\(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{100}}\)) +1
b. B=\(\dfrac{4}{3}-\dfrac{4}{3^2}+...-\dfrac{4}{3^{100}}\)
Chứng minh rằng:
1) B dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{19}1
2) Adfrac{1}{5}+dfrac{2}{5^2}+dfrac{3}{5^3}+dfrac{4}{5^4}+...+dfrac{500}{5^{500}}100
3) Cdfrac{1}{2^3}+dfrac{1}{3^3}+dfrac{1}{4^3}+dfrac{1}{5^3}+...+dfrac{1}{500^3}dfrac{1}{4}
4) Ddfrac{4}{3}+dfrac{10}{9}+dfrac{28}{27}+...+dfrac{3^{98}+1}{3^{98}}100
Làm giúp mình sớm nha! Thanks.
Đọc tiếp
Chứng minh rằng:
1) B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}>1\)
2) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}<100\)
3) \(C=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}<\dfrac{1}{4}\)
4) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}<100\)
Làm giúp mình sớm nha! Thanks.
Bài 1:Tìm x biết:
a)dfrac{6}{5}-2 |1- 3x| 1dfrac{2}{3}
b) (2,8x +50) : dfrac{-3}{2} 51
c) dfrac{x-2}{-2}dfrac{x+4}{3}
d) 4(3-2x) -5(x-1)12
Bài 2: Chứng minh:
dfrac{1}{3^2}+dfrac{1}{4^2}+dfrac{1}{5^2}+...+dfrac{1}{100^2} dfrac{1}{2}
Đọc tiếp
Bài 1:Tìm x biết:
a)\(\dfrac{6}{5}\)-2 |1- 3x| =1\(\dfrac{2}{3}\)
b) (2,8x +50) : \(\dfrac{-3}{2}\) =51
c) \(\dfrac{x-2}{-2}\)=\(\dfrac{x+4}{3}\)
d) 4(3-2x) -5(x-1)=12
Bài 2: Chứng minh:
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
Cho A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Chứng tỏ \(\dfrac{8}{9}>A>\dfrac{2}{5}\)
Chứng minh rằng
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)= \(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)