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sonja333
@sonja333
September 2021
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Помогите, пожалуйста! 1) Вычислите: а) 5¯² б) 12*3¯³ в) (27*3¯²)¯¹ г) 2¯³ * 2²/2¯⁴
2) Упростите выражение и приведите его к виду, не содержащему отрицательных показателей степеней:
а) х¯³ *х^5:х^-6 б) 12у^-7:(3/4у^-5)
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Annatet
1). а) 5^(-2) = (1/5) ^2 = 1/25 б) 12* 3^ (-3) =12* (1/3)^3= 12* 1/27 =4/9 в) (27*3^(-2))^ (-1) = (27* 1/9)^ (-1) = 3^(-1) = 1/3
г) (2^(-3))* ((2^2)/(2^(-4)))= (2^2)/(2^(-1))= 4/1/2= 8
2). a) ((x^(-3))* (x^5))/ (x^(-6)) =(x^5)/(x^(-3))= (x^5)/1/(x^3) = (x^5)*(x^3) = x^8
б) (12y^(-7))/(3/(4y^(-5))) = ((12y^(-7))*(4y^(-5)))/3 = (4y^(-7))*(4y^(-5)) = 16y^(-12)= 16/ y^12
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Answers & Comments
2). a) ((x^(-3))* (x^5))/ (x^(-6)) =(x^5)/(x^(-3))= (x^5)/1/(x^3) = (x^5)*(x^3) = x^8
б) (12y^(-7))/(3/(4y^(-5))) = ((12y^(-7))*(4y^(-5)))/3 = (4y^(-7))*(4y^(-5)) = 16y^(-12)= 16/ y^12