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
K
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19 tháng 3 2018
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
I
30 tháng 3 2020
cách này mình tự nghĩ
\(\hept{\begin{cases}A=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\\B=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\end{cases}}\)
\(\Rightarrow A-B=\left(\frac{4}{7}-\frac{4}{7}\right)+\left(\frac{5}{7^3}-\frac{5}{7^3}\right)+\left(5-5\right)+\left(\frac{3}{7^2}-\frac{6}{7^2}\right)+\left(\frac{6}{7^4}-\frac{5}{7^4}\right)\)
\(\Rightarrow A-B=-\frac{3}{7^2}+\frac{1}{7^4}\)
\(\Rightarrow A-B=\frac{-3\times7^2}{7^4}+\frac{1}{7^4}\)
mà \(-3\times7^2< 1\Rightarrow\frac{1}{7^4}>\frac{-3\times7^2}{7^4}\Rightarrow B>A\)
11 tháng 7 2017
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(-\frac{5}{6}\right)-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{7}{8}+\frac{6}{7}\)
\(=\frac{7}{8}+\frac{6}{7}=\frac{49}{56}+\frac{48}{56}=\frac{49+48}{56}=\frac{97}{56}\)
KK
29 tháng 7 2019
\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
29 tháng 7 2019
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
23 tháng 7 2020
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: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)
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.