So sánh:
A=\(\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\) và B=\(\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
Những câu hỏi liên quan
1. So sánh A và B:
\(A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\); \(B=\frac{5}{7^3}+\frac{6}{7^2}+\frac{5}{7^4}+\frac{4}{7}+5\)
So sánh:frac{frac{frac{1}{2}}{frac{3}{4}}}{frac{frac{5}{6}}{frac{7}{8}}}+frac{frac{frac{8}{7}}{frac{6}{5}}}{frac{frac{4}{3}}{frac{2}{1}}} vàfrac{frac{frac{1}{2}}{frac{3}{4}}+frac{frac{8}{7}}{frac{6}{5}}}{frac{frac{5}{6}}{frac{7}{8}}+frac{frac{4}{3}}{frac{2}{1}}}và frac{frac{frac{1}{2}+frac{8}{7}}{frac{3}{4}+frac{6}{5}}}{frac{frac{5}{6}+frac{4}{3}}{frac{7}{8}+frac{2}{1}}}vàfrac{frac{frac{1+8}{2+7}}{frac{3+6}{4+5}}}{frac{5+4}{frac{6+3}{2+1}}}
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So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
So sánh:
P=\(\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\) và \(Q=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
Giải hộ mik với,đầy đủ+đúng=tick
So sánh:
\(P=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\)
\(Q=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
Ta có : \(P=\left\{\frac{4}{7}+5+\frac{5}{7^3}\right\}+\left\{\frac{3}{7^2}+\frac{6}{7^4}\right\}\)
\(Q=\left\{\frac{4}{7}+5+\frac{5}{7^3}\right\}+\left\{\frac{5}{7^4}+\frac{6}{7^2}\right\}\)
So sánh : \(\frac{3}{7^2}+\frac{6}{7^4}\)và \(\frac{5}{7^4}+\frac{6}{7^2}\)
Ta có : \(\frac{3}{7^2}+\frac{6}{7^4}=\frac{49.3}{7^4}+\frac{6}{7^4}\)
\(\frac{5}{7^4}+\frac{6}{7^2}=\frac{5}{7^4}+\frac{49.6}{7^4}\)
Vì 49.3 + 6 < 49.6 + 5 nên Q > P.
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Bài 1: Tính(hợp lý nếu có thể) a) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\) b) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\) c) \(\frac{-3}{5}.\frac{5}{7}+\frac{-3}{5}.\frac{3}{7}+\frac{-3}{5}.\frac{6}{7}\) d) \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}\)
Bài 1:
a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=4\frac{5}{7}-1\frac{3}{4}\)
\(=\frac{33}{7}-\frac{7}{4}\)
\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)
b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=4\frac{5}{9}-2\frac{3}{4}\)
\(=\frac{41}{9}-\frac{11}{4}\)
\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)
c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)
d) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
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\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(\frac{-5}{6}\right)-\frac{6}{7}-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(-\frac{5}{6}\right)-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{7}{8}+\frac{6}{7}\)
\(=\frac{7}{8}+\frac{6}{7}=\frac{49}{56}+\frac{48}{56}=\frac{49+48}{56}=\frac{97}{56}\)
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Cho:
\(A=4+\frac{1}{7^6}+\frac{3}{7}+\frac{4}{7^2}+\frac{-441}{7^6}+\frac{27}{7^5}\)
\(B=\frac{147}{7}+4+\frac{35}{7^7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
Hãy so sánh A với B
Cho C= \(4+\frac{1}{7^6}+\frac{3}{7}+\frac{4}{7^2}+\frac{-441}{7^6}+\frac{27}{7^5}\)
D=\(\frac{147}{7^3}+4+\frac{35}{7^7}+\frac{4}{7^2}+\frac{27}{7^5}+\frac{-9}{7^9}\)
So sánh C với D
thực hiện phép tính :a, frac{0,4-frac{2}{9}+frac{2}{11}}{1,4-frac{7}{9}+frac{7}{11}}-frac{frac{1}{3}-0,25+frac{1}{5}}{frac{7}{6}-58+5+0,7}b, frac{frac{1}{9}-frac{1}{7}-frac{1}{11}}{frac{4}{9}-frac{4}{7}-frac{4}{11}}+frac{frac{3}{5}-frac{3}{25}-frac{3}{125}-frac{3}{625}}{frac{4}{5}-frac{4}{25}-frac{4}{125}-frac{4}{625}}c, frac{frac{3}{7}-frac{3}{17}+frac{3}{37}}{frac{5}{7}-frac{5}{17}+frac{5}{37}}+frac{frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}}{frac{7}{5}-frac{7}{4}+frac{7}{3}-frac{7}{2}}M...
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thực hiện phép tính :
a, \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{6}-58+5+0,7}\)
b, \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
c, \(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
Mong các bạn giúp đỡ nhé
Tính nhanh: \(\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!}+\frac{5}{3!+4!+5!}+\frac{6}{4!+5!+6!}+\frac{7}{5!+6!+7!}+\frac{8}{6!+7!+8!}\)
Đặt P = ... ( biểu thức đề bài )
Nhận xét: Với \(k\inℕ^∗\) ta có:
\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)
\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)
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