a, Tự chép đề bài ((:

\(=\frac{1}{9}\cdot1+\left(-\frac{1}{243}\right)\cdot\frac{9}{2}\)

\(=\frac{1}{9}-\frac{1}{54}\)

\(=\frac{5}{54}\)

b, 1. \(\left(\frac{2^2\cdot2^3}{4^2\cdot16}\right)^{15}\)

\(=\left(\frac{2^5}{2^4\cdot2^4}\right)^5=\left(\frac{2^5}{2^8}\right)^5=\left(\frac{1}{2^3}\right)^5=\left(\frac{1}{8}\right)^5=\frac{1}{8^5}\)(Để vậy đi :v)

     2. \(\left(\frac{2^6}{16^2}\right)^{10}\)

\(=\left(\frac{2^6}{2^8}\right)^{10}=\left(\frac{1}{2^2}\right)^{10}=\frac{1}{2^{20}}\)

c, \(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}\)

\(=\frac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\frac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\frac{3^2}{1}=3^2=9\)

BÀI 1.

\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\frac{4^5}{5^5}}{\frac{2^6}{5^6}}=\frac{4^5}{5^5}:\frac{2^6}{5^6}=\frac{4^5}{5^5}\cdot\frac{5^6}{2^6}=\frac{4^5\cdot5^6}{5^5\cdot2^6}=\frac{4^5\cdot5}{2^6}=\frac{\left(2^2\right)^5\cdot5}{2^6}=\frac{2^{10}\cdot5}{2^6}\) \(=2^4\cdot5=16\cdot5=80\)

BÀI 2.

\(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{2^{15}\cdot\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\frac{3^8}{3^6}=\frac{3^6\cdot3^2}{3^6}=3^2=9\)

mày muốn gì sao mày ra đề khó thế lần sau ra đề dễ hơn đấy

bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó

đụ cha mi

mi trù ta thi rớt HK II mà ta giúp mày hả

mấy bài này cũng dễ ẹt nữa

đừng có mơ ta sẽ giúp mày

ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha ha
 

\(B=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{99\cdot101}\right)\)

\(B=\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\frac{4^2}{3\cdot5}\cdot\cdot\cdot\frac{100^2}{99\cdot101}\)

\(B=\frac{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot\cdot\cdot99\cdot101}\)

\(B=\frac{\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot100\right)}{\left(1\cdot2\cdot3\cdot\cdot\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot\cdot\cdot101\right)}\)

\(B=\frac{100\cdot2}{1\cdot101}\)

\(B=\frac{200}{101}\)

\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)

\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)

\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)

\(=\frac{2^9.3^9.-17}{1}\)

Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)

\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)

\(=\frac{3.2^{18}}{2^{35}.3^2}\)

\(=\frac{1}{2^{17}.3}\)

Cảm ơn bạn nhé:)