\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)

\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)

\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)

\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)

Bạn xem lại đề

\(125.5^2.\dfrac{1}{625}.5^3=5^3.5^2.\dfrac{1}{5^4}.5^3=5^{3+2-4+3}=5^4\\ 8.32.\left(2^4.\dfrac{1}{32}\right)=2^3.2^5.2^4.\dfrac{1}{2^5}=2^{3+5+4-5}=2^7\\ 6^3.5^2.\left(\dfrac{5}{6}\right)^3=6^3.5^2.5^3:6^3=5^{2+3}.6^{3-3}=5^5.6^0=5^5.1=5^5\\ Bài.5A\)

\(Bài.5B\\ a,2401.\left(\dfrac{1}{7}\right)^2.\dfrac{1}{7}.49^2=7^4.\left(\dfrac{1}{7}\right)^3.\left(7^2\right)^2=7^4.\dfrac{1}{7^3}.7^4=7^{4-3+4}=7^5\\ b,9.81:\left(3^5.\dfrac{1}{27}\right)=3^2.3^4:\left(3^5.\dfrac{1}{3^3}\right)=3^{2+4}:\left(3^{5-3}\right)=3^6:3^2=3^{6-2}=3^4\\ c,3^4.7^2.\left(\dfrac{7}{3}\right)^4=3^4.7^2.7^4:3^4=\left(3^4:3^4\right).\left(7^2.7^4\right)=1.7^6=7^6\)

\(x< 100\Leftrightarrow x\le99\Leftrightarrow x-2\le97\)

 Ta thấy \(\left(x-2\right)⋮5,\left(x-2\right)⋮14\) mà \(\left(5,14\right)=1\) nên \(\left(x-2\right)⋮14.5=70\). Mà \(x-2\le97\) nên \(\left(x-2\right)\in\left\{0;70\right\}\) \(\Leftrightarrow x\in\left\{2;72\right\}\) . 

 Vậy \(x=2\) và \(x=72\) là các số thỏa ycbt.

6³x5²x(\(\dfrac{5}{6}\))³=5²x5³=5⁵.

Cj bé iu ở đây nè cj

30x30x30-100-30=26870.

(\(\dfrac{1}{9}\))² = (\(\dfrac{1^2}{3^2}\))²= ((\(\dfrac{1}{3}\))²)²=  (\(\dfrac{1}{3}\))⁴

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