K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

19 tháng 4 2017

(5/17-5/17)-(5/29-5/29)-(5/37-5/37)-(15-5)=0-0-0-10=-10

31 tháng 3 2016

\(\frac{1774}{145}\)

\(\frac{1774}{145}\)

18 tháng 3 2018

\(\frac{5}{23}x\frac{17}{6}+\frac{5}{23}x\frac{9}{26}\)

\(=\frac{5}{23}x\left(\frac{17}{26}+\frac{9}{26}\right)\)

\(=\frac{5}{23}x1=\frac{5}{23}\)

21 tháng 9 2017

\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)

\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)

\(=\frac{3}{5}+\frac{1}{-7}\)

\(=\frac{16}{35}\)

c) G = \(\frac{636363.37-373737.63}{1+2+3+...+2017}\)

G = \(\frac{63.10101.37-37.10101.63}{1+2+3+...+2017}\)

G = \(\frac{0}{1+2+3+...+2017}\)

=> G = 0

Vậy G = 0

21 tháng 1 2017

a) \(E=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{48.49.50}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(\Rightarrow E=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(\Rightarrow E=\frac{1}{2}.\frac{612}{1225}\)

\(\Rightarrow E=\frac{306}{1225}\)

Vậy...

b) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}=\frac{2.1}{1}=2\)

d) Bạn xem lại đề nhé

3 tháng 8 2019

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\left(-\sqrt{7}\right)+\left(-\sqrt{5}\right)\right)\cdot\frac{\sqrt{7}-\sqrt{7}}{1}\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\cdot\frac{\sqrt{7}-\sqrt{5}}{1}\)

\(=\frac{-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{1}\)

\(=\frac{-\left(7-5\right)}{1}=-2\)