Question

Помогите решить задания по математике

Answer 1

Ответ:

$1)\ \ 27^{-2/3}+32^{3/5}-3^{-2}+5^0=\dfrac{1}{\sqrt[3]{27^2}}+\sqrt[5]{32^3}-\dfrac{1}{3^2}+1=\\\\\\=\dfrac{1}{3^2}+2^3-\dfrac{1}{3^2}+1=2^3+1=8+1=9$

$2)\ \ \dfrac{x+8}{2x+5}=1\ \ \ ,\ \ \ 2x+5\ne 0\ \ \to \ \ x\ne -2,5\ \ ,\\\\\\\dfrac{x+8-2x-5}{2x+5}=0\ \ \ ,\ \ \ \dfrac{-x+3}{2x+5}=0\ \ \Rightarrow \\\\\\-x+3=0\ \ ,\ \ x\ne -2,5\\\\x=3\ \ -\ \ otvet$

$3)\ \ 1\dfrac{3}{5}\cdot \Big(1\dfrac{5}{6}-\dfrac{1}{3}\Big)=\dfrac{8}{5}\cdot \Big(\dfrac{11}{6}-\dfrac{1}{3}\Big)=\dfrac{8}{5}\cdot \dfrac{9}{6}=\dfrac{8}{5}\cdot \dfrac{3}{2}=\dfrac{4\cdot 3}{5}=\dfrac{12}{5}=2,4$