Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)
Thực hiện phép tính ( bằng cách hợp lí nếu có thể )
A. 2/3 - (- 1/4)+3/5-(+7/45)-(-5/9)+1/12+1/39
B. 6.(-2/3)^2 - 3(-2/3)^2 - 2 : (-3/2)+4
A= \(\dfrac{11}{9}\).\(\dfrac{-3}{2}\)- \(\dfrac{11}{9}\).\(\dfrac{15}{2}\)+(2021)0
= \(\dfrac{11}{9}\)(\(\dfrac{-3}{2}\)-\(\dfrac{15}{2}\)) + 1
= \(\dfrac{11}{9}\)(-9) + 1
= -11 + 1
= -10
Lần sau bn ko đc phép chuyển \(\left(-2021\right)^0\) sang \(\left(2021\right)^0\) đou nha
\(\Rightarrow A=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(\Rightarrow A=1+1+\frac{1}{2}\)
\(\Rightarrow A=\frac{5}{2}\)
tíc mình nha
8 − 9 4 + 2 7 − − 6 − 3 7 + 5 4 − 3 + 2 4 − 9 7 = 8 − 9 4 + 2 7 + 6 + 3 7 − 5 4 − 3 − 2 4 + 9 7 = 8 + 6 − 3 − 9 4 − 5 4 − 2 4 + 2 7 + 3 7 + 9 7 = 11 − 4 + 2 = 9
7 3 + − 5 6 + − 2 3 = 7 3 + − 5 6 + − 2 3 = 7 3 + − 2 3 + − 5 6 = 5 3 + − 5 6 = 10 + − 5 6 = 5 6
=3^6 .3^6 /3^12
=3^12/3^12=1
\(\frac{3^6.3^6}{\left(3.27\right)^3}\)
\(\frac{3^6.3^6}{3^3.3^9}\)
\(=\frac{3^{12}}{3^{12}}=1\)