\(\dfrac{8^2.6^3}{9^2.16^2}=\dfrac{\left(2^3\right)^2.2^3.3^3}{\left(3^2\right)^2.\left(2^4\right)^2}=\dfrac{2^{3.2+3}.3^3}{3^4.2^8}=\dfrac{3^3.2^8.2}{3.3^3.2^8}=\dfrac{2}{3}\\ ---\\ \dfrac{\left(0,15\right)^4}{\left(0,5\right)^5}=\left(\dfrac{0,15}{0,5}\right)^4.\dfrac{1}{0,5}=\left(\dfrac{3}{10}\right)^4.2=\dfrac{81}{10000}.2=\dfrac{81}{5000}\\ ---\\ d,\left(\dfrac{3}{4}\right)^3.\left(\dfrac{16}{9}\right)^3=\left(\dfrac{3}{4}.\dfrac{16}{9}\right)^3=\left(\dfrac{48}{32}\right)^3=\left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}\)

b) \(\dfrac{8^2.6^3}{9^2.16^2}=\dfrac{2^6.2^3.3^3}{3^4.2^8}=\dfrac{2^9.3^3}{3^4.2^8}=\dfrac{2}{3}\)

c) \(\dfrac{\left(0,15\right)^4}{\left(0,5\right)^5}=\dfrac{\left(0,5\right)^4.\left(0,3\right)^4}{\left(0,5\right)^5}=\dfrac{0,3^4}{0,5}\)

d) \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{16}{9}\right)^3=\dfrac{3^3}{4^3}.\dfrac{4^6}{3^6}=\dfrac{4^3}{3^3}=\left(\dfrac{4}{3}\right)^3\)

\(\dfrac{5^6.25^3}{25}=\dfrac{5^6.5^6}{5^2}=5^{6+6-2}=5^{10}\)

\(\dfrac{5^6.25^3}{25}=\dfrac{5^6.\left(5^2\right)^3}{5^2}=\dfrac{5^6.5^6}{5^2}=\dfrac{5^{12}}{5^2}=5^{12-2}=5^{10}\)

Chữ số 0 em nhé

Là chữ số 0

giúp mình với

☹

\(a,\widehat{aOn}+\widehat{mOn}+\widehat{mOb}=180^o\left(kề.bù\right)\\ \Leftrightarrow70^o+\widehat{mOn}+40^o=180^o\\ Vậy:\widehat{mOn}=180^o-\left(70^o+40^o\right)=70^o\\ b,Vì:\widehat{aOn}=\widehat{mOn}\\ Mà.tia.On.nằm.giữa.2.tia.Oa.và.Om.nên:\\ On.là.tia.phân.giác.góc.\widehat{aOm}\)

\(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)

        \(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) 

=>      \(\left(2x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\) 

=>        \(2x+\dfrac{1}{2}=\pm\dfrac{1}{2}\) 

TH1: 

\(2x+\dfrac{1}{2}=\dfrac{1}{2}\)

        \(2x=\dfrac{1}{2}-\dfrac{1}{2}=0\) 

          \(x=0\) 

TH2:

\(2x+\dfrac{1}{2}=-\dfrac{1}{2}\)

       \(2x=-\dfrac{1}{2}-\dfrac{1}{2}\)  

        \(2x=-1\) 

            \(x=\dfrac{-1}{2}\) 

      Vậy \(x\in\left\{0;\dfrac{-1}{2}\right\}\)

a) Số hạng thứ 20 (n=20) là 

\(\left(20-1\right).4=76\)

\(A=1-5+9-13+17-21+...+76\)

\(A=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(A=\left(-4\right).38=-152\)

b) Số hạng thứ n là:

\(\left(n-1\right).4\)

\(\)\(A=1-5+9-13+17-21+...+\left(n-1\right).4\)

\(A=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)   ((n-1).2 số -4)

\(A=\left(-4\right).\left(n-1\right).2=-8\left(n-1\right)\)

điên à

\(\dfrac{5}{7}:x+1=\dfrac{-15}{30}\)

\(\dfrac{5}{7}:x=\dfrac{-15}{30}-1\)

\(\dfrac{5}{7}:x=\dfrac{-45}{30}\)

\(x=\dfrac{5}{7}:\dfrac{-45}{30}\)

\(x=\dfrac{-10}{21}\)

\(\dfrac{5}{7}:x+1=-\dfrac{15}{30}\)

\(\Rightarrow\dfrac{5}{7}:x+1=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{1}{2}-1\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{5}{7}:-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{10}{21}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{9}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=3\)

\(\Rightarrow x=3+\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{7}{2}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\\ \\ \\ \Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}=\dfrac{9}{3}=3\\ \\ \\ \Rightarrow x=3+\dfrac{1}{2}=3\dfrac{1}{2}\)