Question

помогите пожалуйста!!!​

Answer 1

$\boxed{(a-b)^2=a^2-2ab+b^2}\\\\\\0{,}3x-\dfrac{5}{x}=6\;\Rightarrow\;\bigg(0{,}3x-\dfrac{5}{x}\bigg)^2=6^2\;\Rightarrow\;(0{,}3x)^2-2\cdot0{,}3x\cdot\dfrac{5}{x}+\bigg(\dfrac{5}{x}\bigg)^2=36\;\Rightarrow\\\\\\\Rightarrow0{,}09x^2-3+\dfrac{25}{x^2}=36\;\Rightarrow\;0{,}09x^2+\dfrac{25}{x^2}=36+3=\boxed{39}$

Answer 2

Ответ:

0,3x−

x

5

=6⇒(0,3x−

x

5

)

2

=6

2

⇒(0,3x)

2

−2⋅0,3x⋅

x

5

+(

x

5

)

2

=36⇒

⇒0,09x

2

−3+

x

2

25

=36⇒0,09x

2

+

x

2

25

=36+3=

39