\(A=\dfrac{\left(-3\right)^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)^{45}\cdot\left(-2\right)^{12}}{5\cdot\left(-3\right)^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)\cdot\left(-2\right)}{5}=\dfrac{6}{5}\)

thi tự làm nhé bạn

Câu 1:

\(a,\text{Slt: }\widehat{A_4}\text{ và }\widehat{B_1}\\ \text{Đv: }\widehat{A_4}\text{ và }\widehat{B_3}\\ b,a\text{//}b\Rightarrow\left\{{}\begin{matrix}\widehat{A_4}=\widehat{B_1}=42^0\left(\text{so le trong}\right)\\\widehat{B_2}=180^0-\widehat{A_4}=138^0\left(\text{trong cùng phía}\right)\end{matrix}\right.\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\\\dfrac{a}{c}=\dfrac{b}{d}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(\dfrac{a}{c}\right)^2=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\\\left(\dfrac{a}{c}\right)^2=\dfrac{ab}{cd}\end{matrix}\right.\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)

\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)

\(x=\)-\(\dfrac{27}{10}\)

\(x:\dfrac{2}{9}=\dfrac{9}{2}\)

\(x.\dfrac{9}{2}=\dfrac{9}{2}\)

\(x=\dfrac{9}{2}:\dfrac{9}{2}\)

\(x=1\)

\(\dfrac{x}{9}=\dfrac{5}{3}\)

\(x.3=5.9\)

\(x.3=45\)

\(x=45:3=15\)

\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)

\(x:\dfrac{8}{125}=\dfrac{125}{8}\)

\(x.\dfrac{125}{8}=\dfrac{125}{8}\)

\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)

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