Tính hợp lý
\(A=\frac{\text{( 2^2. 3 )^6 + 8^4 .3^5 }}{2^{12}.3^5-4^6.9^2}\)
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3 tháng 3 2023
\(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\\ =\dfrac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot\left(3^2\right)^2}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-\left(5^2\right)^5\cdot\left(7^2\right)^2}{\left(5^3\cdot7\right)^3+5^9\cdot\left(2\cdot7\right)^3}\\ =\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\\ =\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^6}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\\ =\dfrac{2}{9}-\dfrac{-6}{1+8}=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
VT
3 tháng 3 2023
\(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\\ =\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\\ =\dfrac{2^{12}.3^5.\left(1-3\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\\ =\dfrac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^6}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\\ =\dfrac{-2}{3}-\dfrac{5.\left(-6\right)}{9}\\ =\dfrac{-2}{3}+\dfrac{30}{9}\\ =\dfrac{8}{3}\)
TA
4 tháng 9 2020
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
4 tháng 9 2020
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
D
19 tháng 5 2017
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
TP
22 tháng 7 2017
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
Phân tích từ số:
\(\frac{2^{12}.3^5-4^2.4^4.3^4}{2^{12}.3.3^5+4^2.4^4.3.3^4}=\frac{1}{6}\)
\(\frac{5^9.5.7^3-5^9.5.7^3.7}{5^9.7^3+5^9.2^3.7^3}=\frac{-10}{3}\)
Sau khi rút gọn là:
\(\frac{1}{6}-\left(-\frac{10}{3}\right)=\frac{1}{6}+\frac{10}{3}=\frac{7}{2}\)
30 tháng 3 2018
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= \(\frac{1}{2}\)
20 tháng 5 2017
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 :)
Ta có : A = \(\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
= \(\frac{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
= \(\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
= \(\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
= \(\frac{3^6+3^5}{3^5-3^4}\)
= \(6\)
\(A=\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
\(A=\frac{2^{12}.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
\(A=\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
\(A=\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
\(\Rightarrow A=\frac{3^5.\left(3+1\right)}{3^4.\left(3-1\right)}=\frac{3^5.4}{3^4.2}=\frac{3.3^4.2.2}{3^4.2}=\frac{3.2}{1}=6\)