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K
Khách
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CB
2
KN
7 tháng 8 2020
a. \(\frac{7}{8}< \frac{x}{35}< \frac{15}{7}\)
\(\Rightarrow\frac{245}{280}< \frac{8x}{280}< \frac{600}{280}\)
\(\Rightarrow245< 8x< 600\)
\(\Rightarrow30< x< 75\)
\(\Rightarrow x\in\left\{31;32;33;...;72;73;74\right\}\)
b. \(\frac{21}{3}< \frac{x}{7}\le\frac{24}{2}\)
\(\Rightarrow\frac{294}{42}< \frac{6x}{42}\le\frac{504}{42}\)
\(\Rightarrow294< 6x\le504\)
\(\Rightarrow49< x\le84\)
\(\Rightarrow x\in\left\{50;51;52;...;82;83;84\right\}\)
16 tháng 6 2016
*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)
*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
21 tháng 7 2018
\(A=\frac{14^{16}-21^{32}.35^{68}}{10^{16}.15^2.7^{96}}=\frac{2^{16}.7^{16}-3^{32}.7^{32}.7^{68}.5^{68}}{5^{16}.2^{16}.3^2.5^2.7^{96}}\)
\(A=\frac{2^{16}.7^{16}-3^{32}.7^{100}.5^{68}}{5^{18}.2^{16}.3^2.7^{96}}=\frac{7^{16}.\left(2^{16}-3^{32}.7^{84}.5^{68}\right)}{5^{18}.2^{16}.3^2.7^{96}}=\frac{2^{16}-3^{32}.7^{84}.5^{68}}{5^{18}.2^{16}.3^2.7^{88}}\)
\(B=\frac{4^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}=\frac{2^{40}-2^{20}+2^{20}.3^{20}}{2^{20}.3^{20}-3^{20}+3^{40}}=\frac{2^{20}.\left(2^{20}-1+3^{20}\right)}{3^{20}.\left(2^{20}-1+3^{20}\right)}\)
\(B=\frac{2^{20}}{3^{20}}\)
NT
0
Mình làm câu này các câu khác tương tự
Mấu chốt làm bài này là đưa về thừa số nguyên tố
\(\frac{12^7.3^{35}.14^{21}}{9^{21}.7^{21}.2^{35}}=\frac{3^7.4^7.3^{35}.2^{21}.7^{21}}{3^{2.21}.7^{21}.2^{35}}=\frac{3^7.2^{2.7}.3^{35}.2^{21}.7^{21}}{3^{2.21}.7^{21}.2^{35}}=\frac{3^7.2^{14}.3^{35}.2^{21}.7^{21}}{3^{42}.7^{21}.2^{35}}=\frac{3^{42}.2^{35}.7^{21}}{3^{42}.7^{21}.2^{35}}=1\)