\(\frac{16^33^{10}+144\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)
rút gọn p/s trên
giúp MEOWW ÙwÚ
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\(\dfrac{16^3.3^{10}+144.6^9}{4^6.3^{12}+6^{11}}\)
=\(\dfrac{16^3.3^{10}+144.6^9}{4.4.4.4.4.4.3^{12}+6^{11}}\)
=\(\dfrac{16^3.3^{10}+144.6^9}{16.16.16.3^{12}+6^{11}}\)
=\(\dfrac{16^3.3^{10}+4.6^2.6^9}{16^3.3^{12}+6^{11}}\)
=\(\dfrac{4.6^{11}}{3^2+6^{11}}\)
=\(\dfrac{4}{9}\)
Ko chắc
tử số 46.310+144.69 = 212.310+2.23.3.3.69
= 69.(23.3)+(23.3).69.6 = 69.(23.3).(6+1) = 69.(23.3).7 = 69.2.3.22.7
= 610.4.7
mẫu số 46.312+611 = 212.312+611 = 612+611 = 611.6+611
= 611.(6+1) = 611.7
\(\dfrac{16^{3}.3^{10}+144.6^{9}}{4^{6}.3^{12}.6^{11}}=\dfrac{6^{10}.4.7}{6^{11}.7}\)=\(\dfrac{4}{6}=\dfrac{2}{3}\)
mà cậu trình bày trên phân số nhé để nhanh thì mình chia ra thôi
Ta có: \(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(6+1\right)}\)
\(=\frac{2.6}{3.7}=\frac{4}{7}\)
\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{2^{12}.3^{10}+120.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^3.2^9.3^9.3+120.2^9.3^9}{2.2^{11}.3^{11}.3+2^{11}.3^{11}}\)
\(=\frac{2^9.3^9.\left(2^3.3+120\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2^9.3^9.144}{2^{11}.3^{11}.7}\)
\(=\frac{2^9.3^9.2^4.3^2}{2^{11}.3^{11}.7}\)
\(=\frac{2^2.2^7.3^9.2^4.3^2}{2^{11}.3^{11}.7}\)
\(=\frac{4.2^{11}.3^{11}}{2^{11}.3^{11}.7}=\frac{4}{7}\)
a) \(3^{30}\) và \(5^{20}.\)
Ta có:
\(3^{30}=\left(3^3\right)^{10}=27^{10}.\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}.\)
Vì \(27>25\) nên \(27^{10}>25^{10}.\)
\(\Rightarrow3^{30}>5^{20}.\)
Chúc bạn học tốt!
a/ Có: \(3^{30}=\left(3^3\right)^{10}=27^{10}\\ 5^{20}=\left(5^2\right)^{10}=25^{10}\)
Mà \(27^{10}>25^{10}\)
\(\Rightarrow3^{30}=5^{20}\)
b/ \(A=\frac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\\ A=\frac{\left(2^4\right)^3\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{\left(2^2\right)^6\cdot3^{12}+2^{11}\cdot3^{11}}\\ A=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\\ A=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\\ A=\frac{2\cdot6}{3\cdot7}=\frac{4}{7}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=-\frac{1}{3}\)
Ta có:\
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(A=-\frac{2}{6}=-\frac{1}{3}\)
\(M=\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(M=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(M=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(M=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(M=\frac{-2}{6}=\frac{-1}{3}\)
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
Bài làm của mk hơi tắt nên bạn tự suy luận nhé
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)=\(\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)=\(\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}\)=\(\frac{-13122}{6561.6}\)=\(-\frac{1}{3}\)
Ta có :
\(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(6+1\right)}=\frac{12}{21}=\frac{4}{7}\)
Chúc bạn học tốt ~
\(\frac{16^3.3^{10}+144.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^4.3^2.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^{13}.3^{11}}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+2.3\right)}{2^{11}.3^{11}.\left(1+2.3\right)}=\frac{2}{3}\)