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Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)
=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)
\(=\frac{5}{2}.\frac{6}{5}-17\)
\(=3-17=-14\)
Tụi quá mới lớp 5 thui
Ta có
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\cdot\frac{5^4.20^4}{25^5.4^5}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\cdot\frac{2^8.5^8}{5^{10}.2^{10}}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\cdot\frac{1}{5^2.2^2}\)
\(=\frac{\left(-2\right)}{6}\cdot\frac{1}{100}=-\frac{1}{3}\cdot\frac{1}{100}=-\frac{1}{300}\)
Vậy : \(E=-\frac{1}{300}\)
Bài làm
\(E=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}.\frac{5^4.20^4}{25^5.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}.\frac{5^4.4^4.5^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\frac{5^8.4^4}{5^{10}.4^5}\)
\(\Rightarrow E=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\left(3^8+3^8.5\right)}.\frac{1}{5^2.4}\)
\(\Rightarrow E=\frac{3^8-3^9}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=\frac{3^8\left(1-3\right)}{3^8\left(1+5\right)}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{2}{6}.\frac{1}{100}\)
\(\Rightarrow E=-\frac{1}{300}\)
thank bạn
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
Câu b) tạm thời ko bít làm =.=
Bài 1 :
\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)
\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)
\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)
\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)
\(\Leftrightarrow\)\(2^{12}=2x\)
\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)
\(\Leftrightarrow\)\(x=2^{11}\)
\(\Leftrightarrow\)\(x=2048\)
Vậy \(x=2048\)
Chúc bạn học tốt ~
Bài 1 :
\(a)\) Ta có :
\(4+\frac{x}{7+y}=\frac{4}{7}\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)
\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)
\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)
Do đó :
\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)
\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)
Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)
Chúc bạn học tốt ~
a)A=\(\frac{3^8.20^5-3^9.5^5.2^9}{6^8.10^4-3^8.2^9.5^4}\)
A=\(\frac{2^{10}.3^8.5^5-2^9.3^9.5^5}{2^{12}.3^8.5^5-2^9.3^9.5^5}\)
A=\(\frac{\left(2-3\right).\left(2^9.3^8.5^5\right)}{\left(2^3-3\right).\left(2^9.3^8.5^5\right)}\)
A=\(\frac{2-3}{8-3}\)
A=\(\frac{-1}{5}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(\frac{3}{5}B=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)
\(\frac{3}{5}B=\frac{7-4}{4.7}+\frac{7-10}{7.10}+\frac{13-10}{10.13}+...+\frac{28-25}{25.28}\)
\(\frac{3}{5}B=\frac{7}{4.7}-\frac{4}{4.7}+\frac{10}{7.10}-\frac{7}{7.10}+\frac{13}{10.13}-\frac{10}{10.13}+...+\frac{28}{25.28}-\frac{25}{25.28}\)
\(\frac{3}{5}B=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{7}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\)
\(\frac{3}{5}B=\frac{1}{4}-\frac{1}{28}\)
\(\frac{3}{5}B=\frac{3}{14}\)
\(B=\frac{5}{14}\)