спросив выраз 0,2 х 12 y -9. 50 х -10 у 10.. Created by Voshina02. matematika-ru

спросив выраз 0,2 х 12 y -9. 50 х -10 у 10....

$\boxed{ {n}^{a} \times {n}^{b} = {n}^{a + b} }$

$0.2 {x}^{12} {y}^{ - 9} \times 50 {x}^{ - 10} {y}^{10} = 0.2 \times 50 \times {x}^{12} \times {x}^{ - 10} \times {y}^{ - 9} \times {y}^{10} = 10 \times {x}^{12 + ( - 10)} \times {y}^{ - 9 + 10} = \boldsymbol{10 {x}^{2} y}$

1 votes Thanks 1

kamilmatematik100504

Ответ: $\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\la\la\la\la\ddddddddddddddddddddddddddddddddcleverdddddd\ffffffffffffffffffffffffffffffffffffffff\pppppppppppppppppppppppppppppppppppp\ddddddddddddddddddd \displaystyle \Large \boldsymbol{} 1) \ 10x^2y \\\\\\2) \ x_{1;2 }=\pm\sqrt{3}$

Пошаговое объяснение:

$\displaystyle\ \Large \boldsymbol{} \huge \boxed{ \ \boxed{a^n\cdot a^{-m}=a^{n-m}} \ } \\\\\\ \Large \boldsymbol{} 1) \ 0,2\ \underline{x}^{12}\cdot \boldsymbol{y}^{-9}\cdot 50 \ \underline{x}^{-10} \cdot \boldsymbol{y}^{10}= \\\\\\ \underbrace{0,2\cdot 50}_{10} \cdot \ \ x^{12-10}\cdot y^{10-9}=\boxed{10x^2y} \\\\\\2) \ x^4+x^2-12 \ \ \ \ \boxed{\ t=x^2 \ ; \ t^2=x^4\ \ ; \ \ t>0 }\\\\\\t^2+t-12=0$ $\displaystyle\Large \boldsymbol{} \left \{ {{t_1+t_2=-1} \atop \!\!\!\!\!{t_1t_2=-12}} \right.=> t_1=-4 \ \o \ ; \ \ \boxed{t_2=3 } \\\\\\x^2=3 \Longrightarrow x_{1; 2} =\pm\sqrt{3}$

2 votes Thanks 2

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