Задали примеры решить.100 1) (4x^2-y)7y: (49x^2-y^2) 2)(7x-y)7: (49x^2-y^2) 3)(4x^2-25y^2)*(1: (2x+5y)+1: (2x-5y)) 4)(2x^2+xy-6y^2): (3y-2x)

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maksimovaa1

Алгебра

4,5(12 оценок)

==========================================================(7x-y)*7y: (49x^2-y^2) = (7x-y)*7y/49x^2-y^2 = (7x-y)*7y/(7x-y)*(7x+y) = 7y/7x+y========================================================== 4x: (7y+-7y): (2x+7y)= 4x/2x+7y - 2x-7y/2x+7y = 4x-(2x-7y)/2x+7y = 4x-2x+7y/2x+7y = 2x+7y/2x+7y = 1 ========================================================== (4x^2-25y^2)*(1: (2x+5y)+1: (2x-5y)) = (2x-5y) * (2x+5y) * (1/2x+5y + 1/ 2x-5y) = (2x-5y) * (2x+5y) * 2x-5y+2x+5y/ (2x+5y)*(2x-5y)= 4x ========================================================== (2x^2+xy-6y^2)/(3y-2x)=2x^2-3xy+4xy-6y^2/3y-2x = -x(3y-2x)-2y(2x-3y-2x)/3y-2x = (3y--2y)/3y-2x = -x-2y==========================================================

dkdjdhhs

Алгебра

4,6(57 оценок)

1) b4=b1•q^3 b2=b1•q b3=b1•q^2 {b1•q^3–b1•q=30 {b1•q^3–b1•q^2=24 {b1•q(q^2–1)=30 {b1•q^2(q–1)=24 {b1•q(q–1)(q+1)=30 {b1•q^2(q–1)=24 разделим первое на второе: q+1 5 = q 4 4(q+1)=5q 4q+4=5q q=4 b1=30/q(q^2–1)=30/60=1/2 ответ: b1=1/2; q=4 2) b2=b1•q b5=b1•q^4 b3=b1•q^2 b4=b1•q^3 {b1•q–b1•q^4=78 {b1•q^2+b1•q^3+b1•q^4=–117 {b1•q(1–q^3)=78 {b1•q^2(1+q+q^2)=–117 разделим первое на второе: 1–q^3 2 = – q(1+q+q^2) 3 (1–q)(1+q+q^2) 2 = – q(1+q+q^2) 3 1–q 2 = – q 3 3–3q = –2q q=3 b1=78/q(1–q^3)=78/3•26=78/78=1 ответ: b1=1; q=3