bn vt lớn dữ z -.-

\(\frac{15^{19}.5^{10}}{9^{10}.5^{14}}=\frac{3^{19}.5^{19}.5^{10}}{3^{20}.5^{14}}=\frac{3^{19}.5^{29}}{3^{20}.5^{14}}=\frac{5^{15}}{3}\)

\(\frac{15^{19}.5^{10}}{9^{10}.5^{14}}=\frac{3^{19}.5^{19}.5^{10}}{3^{20}.5^{14}}=\frac{3^{19}}{3^{20}}.\frac{5^{29}}{5^{14}}\)\(=\frac{1}{3}.\frac{5^{15}}{1}=\frac{5^{15}}{3}\)

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\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)

\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)

\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)

\(\Leftrightarrow X=\frac{109}{6075}\)

Vậy X=109/6075

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Đặt: 

 \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)

=> \(A=\frac{12}{25}\)

Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)

   \(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)

=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)

Giải phương trình:

\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)

                        \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)

                                                                            \(12x+\frac{12}{25}=11x+\frac{121}{243}\)

                                                                             \(12x-11x=\frac{121}{243}-\frac{12}{25}\)

                                                                                                  \(x=\frac{109}{6075}\)

giúp mk với các bạn ơi

\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=5\)

\(\frac{9^{15}.25^{43}}{27^{10}.5^{85}}=\frac{\left(3^2\right)^{15}.\left(5^2\right)^{43}}{\left(3^3\right)^{10}.5^{85}}=\frac{3^{30}.5^{86}}{3^{30}.5^{85}}=1.5=5.\)

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tính thường hay tính nhanh vậy bạn?

Nguyễn Hoàng Thiên Di

tính thường cx đc mà tính nhanh càng tốt nha

đợi mik xíu vì giải hơi dài ạ

\(x=\frac{2^3\cdot25^2}{10^2\cdot5^3}=\frac{2^3\cdot\left(5^2\right)^2}{\left(2\cdot5\right)^2\cdot5^3}\)

    \(=\frac{2^3\cdot5^4}{2^2\cdot5^2\cdot5^3}=\frac{2^3\cdot5^4}{2^2\cdot5^5}=\frac{2}{5}\)

Vậy x = 2/5

x=2^3.25^2/10^2.5^3

x=8.625/100.125

x=5000/12500

x=10/25

x=0,4

\(x=\frac{2^3.25^2}{10^2.5^3}=\frac{2^3.5^4}{2^2.5^2.5^3}=\frac{2^3.5^4}{2^2.5^5}\)\(=\frac{2}{5}\)