\(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.\left(3^4\right)^2}{3^5.3^2}=\frac{3^2.3^8}{3^7}=3^3\)

\(3^6.256^2.2^4=3^6.\left(2^8\right)^2.2^4=3^6.2^{16}.2^4=3^6.2^{20}\)

p/s: chúc bạn học tốt

a)\(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.81^2}{243.3^2}=\frac{3^2.\left(3^4\right)^2}{\left(3^5\right).3^2}=\frac{3^2.3^8}{3^5.3^2}=3^3\)

b)\(3^6.256^2.2^4=3^6.\left(2^8\right)^2.2^4=3^6.2^{16}.2^4=3^6.2^{20}\)

Chúc bạn học tốt

a) \(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{1.81^2}{243}.\frac{3^2}{3^2}=\frac{6561}{243}.1=27\)

b, \(4^6.256^2.2^4=2^{12}.2^{16}.2^4=2^{32}\)

c)  \(A=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)(Mình rút gọn lun cho nhanh nhé ) \(\Rightarrow A=\frac{4}{5}\)

d) \(\Rightarrow B=70\)k cho mình nha Cô Nàng Họ Dương

Đây nhé : ý a,b mình đã giải thích rồi 

c) \(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{2.6}{3.5}=\frac{12}{15}=\frac{4}{5}\)\(\frac{4}{5}\)

d) \(=\frac{2^4.5^4+2^5.5^3}{2^3.5^2}=\frac{2^4.5^3.\left(5+2\right)}{2^3.5^2}=2.5.7=70\)

Đặt: \(B=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+........+2019}\)

Ta có: \(1+2=\frac{2.3}{2}\); \(1+2+3=\frac{3.4}{2}\); .............. ; \(1+2+3+......+2019=\frac{2019.2020}{2}\)

\(\Rightarrow B=\frac{2}{2}+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+........+\frac{1}{\frac{2019.2020}{2}}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{2019.2020}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2019.2020}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2.\left(1-\frac{1}{2020}\right)=2.\frac{2019}{2020}=\frac{2019}{1010}\)

\(\Rightarrow A=\frac{2.2019}{\frac{2019}{1010}}=2.1010=2020\)

viet lai di

bn ko hiểu ak

* là nhân mà . còn ^ là mũ 

a) \(64^2\cdot32^4=2^{16}\cdot2^{20}=2^{36}\)

b) \(11^{16}\cdot5^{24}=\left(11^4\right)^4\cdot\left(5^6\right)^4=\left(11^4\cdot5^6\right)^4\)

a)64².32⁴=(2⁶)².(2⁵)⁴=2¹².2²⁰=2³²

b)11¹⁶.5²⁴(ko phân tích đc nữa)

voa2c0a94e3ca

\(\frac{-11}{9}\le x+\frac{11}{18}\Leftrightarrow x\ge\frac{-11}{6}\)

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow\frac{-13}{9}\le x-\frac{-11}{18}\)

\(\Leftrightarrow\frac{-13}{9}\le x+\frac{11}{18}\)

\(\Rightarrow x\ge\frac{-37}{18}\)

Học tốt nhé !! ^-^

cảm ơn Lâm Việt Phúc nhé

Ta có: \(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}\)

\(=\frac{1}{3}\left(11+2\sqrt{30}\right)-\frac{\sqrt{30}}{2}-\sqrt{30}\)

\(=\frac{11}{3}+\frac{2}{3}\sqrt{30}-\frac{\sqrt{30}}{2}-\sqrt{30}\)

\(=\frac{11}{3}-\frac{5}{6}\sqrt{30}\)

\(=\frac{22-5\sqrt{30}}{6}\)

Ta có: \(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right)\div\sqrt{\frac{81}{6}}\)

\(=\left(\frac{\sqrt{6}}{6}-\frac{9\sqrt{6}}{4}+\frac{2\sqrt{6}}{9}\right)\div\frac{3\sqrt{6}}{2}\)

\(=-\frac{67\sqrt{6}}{36}\cdot\frac{2}{3\sqrt{6}}\)

\(=-\frac{67}{54}\)