Объяснение:
x=0,5⁻¹.
x⁻²=(0,5⁻¹)⁻²=0,5²=0,25.
Пусть х⁻²=t =0,25. ⇒
[tex]\frac{5t}{4-t}-\frac{5t}{4+t}=\frac{5t(4+t)-5t*(4-t)}{(4-t)*(4+t)} =\frac{20t+5t^2-20t+5t^2}{16-t^2} = \frac{10t^2}{16-t^2} =\frac{10*0,25}{16-0,25}=\frac{2,5}{15,75}=\frac{10}{63} .[/tex]
x=0,5⁻¹=(1/2)⁻¹=2.
[tex]OTBET:\ \frac{5x^{-2}}{4-x^{-2}}-\frac{5x^{-2}}{4+x^{-2}}=\frac{10}{63}.\ \ \ \ \ x=2.[/tex]
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Answers & Comments
Объяснение:
x=0,5⁻¹.
x⁻²=(0,5⁻¹)⁻²=0,5²=0,25.
Пусть х⁻²=t =0,25. ⇒
[tex]\frac{5t}{4-t}-\frac{5t}{4+t}=\frac{5t(4+t)-5t*(4-t)}{(4-t)*(4+t)} =\frac{20t+5t^2-20t+5t^2}{16-t^2} = \frac{10t^2}{16-t^2} =\frac{10*0,25}{16-0,25}=\frac{2,5}{15,75}=\frac{10}{63} .[/tex]
x=0,5⁻¹=(1/2)⁻¹=2.
[tex]OTBET:\ \frac{5x^{-2}}{4-x^{-2}}-\frac{5x^{-2}}{4+x^{-2}}=\frac{10}{63}.\ \ \ \ \ x=2.[/tex]