\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}=?\)
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\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
eo ôi t làm rồi mà bị xoá :v thôi t hướng dẫn :v
Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)
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\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{18}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\left(2^6\cdot3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{2}{193}+\dfrac{5\cdot2}{3}=\dfrac{1936}{579}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6.\left(3+1\right)}-\frac{5^9.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3.1\right)}\)
\(=\frac{-1}{2}-\frac{-2}{3}=\frac{-7}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a hơi dài để làm phần b trước :
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\cdot3^2-2^n\cdot2^2+3^n-2^n\)
\(=\left(3^n\cdot3^2+3^n\right)-\left(2^n\cdot2^2+2^n\right)\)
\(=3^n\cdot\left(3^2+1\right)-2^n\cdot\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^3.3\right)^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{18}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5.\left(2^6-1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(A=\frac{2}{3.\left(64-1\right)}-\frac{5.\left(-6\right)}{9}\)
\(A=\frac{2}{3.63}+\frac{30}{9}\)
Tự lm tiếp Ball nhé~
\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^3\cdot3\right)^6+8^4\cdot3^5}=\frac{2}{579}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{18}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^5\left(2^6\cdot3+1\right)}=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot193}=\frac{2}{3\cdot193}=\frac{2}{579}\)