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07.02.2022 01:50
Алгебра
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1. (a+2b)²-(3c+4d)²= 2. (m-2n)²-(2p-3q)²= 3. 9(m+n)²-(m-n)²= 4. 16(a+b)²-9(x+y)²= 5. (x-y)²-(m+n)²= 6. (2a-3c)²-(4b+5d)²= 7. 4(a-b)²-(a+b)²= 8. 9(a-b)²-4(x-y)²=

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Ответы на вопрос:

Mikich2
Mikich2
4,6(27 оценок)
1.  (a+2b)²-(3c+4d)²=a²+4ab+4b²-9c²-24cd-16d²2.  (m-2n)²-(2p-3q)²=m²-4mn+4n²-4p²+12pq-9q² 3. 9(m+n)²-(m-n)²=9(m²+2mn+n²)-m²+2mn-n²=9m²+18mn+9n²-m²+2mn-n²=8m²+20mn+8n² 4. 16(a+b)²-9(x+y)²==16(a²+2ab+b²)-9(x²+2xy+y²)=16a²+32ab+16b²-9x²-18xy-9y² 5. (x-y)²-(m+n)²=x²-2xy+y²-m²-2mn-n²6. (2a-3c)²-(4b+5d)²=4a²-12ac+9c²-16b²-40bd-25d² 7. 4(a-b)²-(a+b)²=4(a²-2ab+b²)-a²-2ab-b²=4a²-8ab+4b²-a²-2ab-b²=3a²-10ab+3b² 8. 9(a-b)²-4(x-y)²=9(a²-2ab+b²)-4(x²-2xy+y²)=9a²-18ab=9b²-4x²+8xy-4y²
123456532
123456532
4,4(55 оценок)

Объяснение:

Уравнение к-ой степени имеет к корней в поле комплексных чисел

cos(π/12)=cos[(π/3)-(π/4)]=cos(π/3)cos(π/4)+sin(π/3)sin(π/4)=

=0,5·(√2/2)+(√3/2)(√2/2)=(√6+√2)/4

sin(π/12)=sin[(π/3)-(π/4)]=sin(π/3)cos(π/4)-cos(π/3)sin(π/4)=

=(√3/2)(√2/2)-0,5·(√2/2)=(√6-√2)/4

x⁶+3i=0

x⁶=-3i=3(cos(-π/2)+isin(-π/2))

x₀= (cos(-π/12)+isin(-π/12))= (cos(π/12)-isin(π/12))=

=((√6+√2)/4-i(√6-√2)/4)=((√6+√2)-i(√6-√2))/4

x₁= (cos((-π+2π)/12)+isin((-π+2π)/12))= (cos((π)/12)+isin((π)/12))=

((√6+√2)+i(√6-√2))/4

x₂= (cos((-π+4π)/12)+isin((-π+4π)/12))= (cos((3π)/12)+isin((3π)/12))=

= (cos((π)/4)+isin((π)/4))= (√2/2+i√2/2)= √2(1+i)/2

x₃= (cos((-π+6π)/12)+isin((-π+6π)/12))= (cos((5π)/12)+isin((5π)/12))=

= (sin((π)/12)+icos((π)/12))= ((√6-√2)/4+i(√6+√2)/4)=

=((√6-√2)+i(√6+√2))/4

x₄= (cos((-π+8π)/12)+isin((-π+8π)/12))= (cos((7π)/12)+isin((7π)/12))=

= (-sin((π)/12)+icos((π)/12))= (-(√6-√2)/4+i(√6+√2)/4)=

=((√2-√6)+i(√6+√2))/4

x₅= (cos((-π+10π)/12)+isin((-π+10π)/12))= (cos((3π)/4)+isin((3π)/4))=

= (-cos((π)/4)+isin((π)/4))= (-√2/2+i√2/2)= √2(-1+i)/2

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