\(\dfrac{-1}{3};\dfrac{7}{9};\dfrac{5}{6}\)So sánh các số hữu tỉ
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
bài này có trong sách Nâng cao và Phát triển bạn nhé
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
b) B=\(\dfrac{\dfrac{1}{22}+\dfrac{1}{13}-0,5}{\dfrac{3}{13}-\dfrac{3}{2}+\dfrac{3}{22}}.\dfrac{\dfrac{3}{4}-0,375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0,875}+\dfrac{3}{4}\)
Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)
\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)
\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)
tìm x,y viết dưới dạng phân số
a. \(5+\dfrac{x}{5+\dfrac{2}{5+\dfrac{3}{5+\dfrac{4}{5}}}}=\dfrac{x}{1+\dfrac{5}{2+\dfrac{4}{3+\dfrac{3}{5+\dfrac{1}{6}}}}}\)
b.
\(\dfrac{y}{3+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{3}}}}}}+\dfrac{y}{7+\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{4}}}}\)
= 2
c,
\(x.\left(\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+1}}}}}}}}\right)=\)\(2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}}}}\)+\(x.\left(3+\dfrac{1}{3-\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{3-\dfrac{1}{3}}}}}\right)\)
Giair bằng máy tính casio
bài này đúng là thị của phi...vô của lí ... :))
8) \(A=\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
9) \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2014}}+\dfrac{1}{3^{2015}}\)
10) \(P=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2005}}{\dfrac{2004}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+...+\dfrac{1}{2004}}\)
8,A=\(\dfrac{9}{10}-\left(\dfrac{1}{10\times9}+\dfrac{1}{9\times8}+\dfrac{1}{8\times7}+...+\dfrac{1}{2\times1}\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+...+\dfrac{1}{2}-1\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-1\right)\)
=\(\dfrac{9}{10}-\dfrac{\left(-9\right)}{10}\)
=\(\dfrac{9}{5}\)
Tính:
a/\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{3+\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{3}{4}}{2-\dfrac{2}{2}+\dfrac{2}{3}-\dfrac{2}{4}}\)
b/\(\dfrac{1+\dfrac{1}{4}+\dfrac{1}{1+\dfrac{1}{4}}}{1-\dfrac{1}{4}-\dfrac{1}{1-\dfrac{1}{4}}}\)
c/\(\dfrac{\dfrac{2}{5}-\dfrac{7}{5}}{\dfrac{2}{5}-\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}.\dfrac{3}{7}-1}}-\dfrac{1}{\dfrac{3}{7}\left(\dfrac{3}{4}.\dfrac{3}{7}.\dfrac{2}{5}-\dfrac{2}{5}-\dfrac{3}{4}\right)}\)
d/\(\left(\dfrac{\dfrac{4}{3}}{2+\dfrac{4}{3}}+\dfrac{2-\dfrac{4}{3}}{\dfrac{4}{3}}\right).\left(\dfrac{\dfrac{2}{3}}{4+\dfrac{2}{3}}-\dfrac{4-\dfrac{2}{3}}{\dfrac{2}{3}}\right)\)
Giúp mik với các bạn ơi 1 bài thôi cug đc.
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
\(3\dfrac{3}{3}.\dfrac{1}{3}-\dfrac{3}{4}.\dfrac{1}{3}\)
\(\left[\dfrac{11}{3}\right]-\left(\dfrac{-1}{2}\right)^2-4\dfrac{1}{2}\)
\(\left(\dfrac{3}{2}-\dfrac{5}{4}+\dfrac{1}{3}\right):\left(\dfrac{4}{3}+2\dfrac{3}{2}-\dfrac{3}{4}\right)\)
\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5+\dfrac{-5}{27}+\dfrac{16}{23}\)
\(2\dfrac{5}{4}+\left(-2018\right)^0-\left[\dfrac{-1}{4}\right]\)
\(\dfrac{19}{11}.\dfrac{6}{5}+\dfrac{6^2}{11}.\dfrac{6}{5}-\left(\dfrac{1}{2}\right)^0\)
\(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\right).3^5+\left(\dfrac{1}{3^5}+\dfrac{1}{3^6}+\dfrac{1}{3^7}+\dfrac{1}{3^8}\right).3^9\)
\(=\dfrac{3^5}{3}+\dfrac{3^5}{3^2}+\dfrac{3^5}{3^3}+\dfrac{3^5}{3^4}+\dfrac{3^9}{3^5}+\dfrac{3^9}{3^6}+\dfrac{3^9}{3^7}+\dfrac{3^9}{3^8}\\ =2\left(3^4+3^3+3^2+3\right)\)
\(=\left(\dfrac{3^5}{3}+\dfrac{3^5}{3^2}+\dfrac{3^5}{3^3}+\dfrac{3^5}{3^4}\right)+\left(\dfrac{3^9}{3^5}+\dfrac{3^9}{3^6}+\dfrac{3^9}{3^5}+\dfrac{3^9}{3^8}\right)\)
\(=\left(3^4+3^3+3^2+3\right)+\left(3^4+3^3+3^2+3\right)\)
\(=2\left(3^4+3^3+3^2+3\right)\)
\(=2\left(81+27+9+3\right)\)
\(=2\left(120\right)\)
\(=240\)
Chứng minh:
a. \(A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
b.\(B=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}< \dfrac{3}{16}\)
c. \(C=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{79}+\dfrac{1}{80}>\dfrac{7}{12}\)
Bài 1.(2,5 điểm)Tìm x, biết:
a) \(\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right).x=-4\dfrac{1}{6}+3\dfrac{1}{2}\)
b) \(\left(1\dfrac{1}{3}+3\dfrac{1}{2}\right).x=4\dfrac{1}{6}-3\dfrac{1}{2}\)
c) \(\dfrac{1}{3}-\dfrac{7}{8}.x=\dfrac{1}{4}\)
d) \(\dfrac{3}{2}.x+\dfrac{1}{7}=\dfrac{7}{8}.\dfrac{64}{49}\)
e) \(5\dfrac{1}{2}-\left(\dfrac{1}{4}.x+\dfrac{2}{5}\right)=25\%\)
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)