Question

помогите пожалуйста прошуууууууу

Answer 1

Ответ:

...

Объяснение:

№1

1)(a+4)^2=a^2+8a+16

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25(x^2+y)(x^2-y)=x^4-y^2

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25(x^2+y)(x^2-y)=x^4-y^22)0.36-c^2=(0.6)^2-c^2=(0.6-c)(0.6+c)

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25(x^2+y)(x^2-y)=x^4-y^22)0.36-c^2=(0.6)^2-c^2=(0.6-c)(0.6+c)a^2+10a+25=(a+5)^2

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25(x^2+y)(x^2-y)=x^4-y^22)0.36-c^2=(0.6)^2-c^2=(0.6-c)(0.6+c)a^2+10a+25=(a+5)^24)3(1+2xy)=(3+6xy)(1-2xy)=3-6xy*2xy=3-12x^2*y^2=-12x^2*y^2+3

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25(x^2+y)(x^2-y)=x^4-y^22)0.36-c^2=(0.6)^2-c^2=(0.6-c)(0.6+c)a^2+10a+25=(a+5)^24)3(1+2xy)=(3+6xy)(1-2xy)=3-6xy*2xy=3-12x^2*y^2=-12x^2*y^2+3(a+6)^2-(a-6)^2=(a+6-(a-6))*(a+6+a-6)=(a+6-a+6)*2a=12*2a=24a

1)(a+4)^2=a^2+8a+16(3y-c)^2=c^2-6cy+9y^2(2a+5)(2a+5)=4a^2-25(x^2+y)(x^2-y)=x^4-y^22)0.36-c^2=(0.6)^2-c^2=(0.6-c)(0.6+c)a^2+10a+25=(a+5)^24)3(1+2xy)=(3+6xy)(1-2xy)=3-6xy*2xy=3-12x^2*y^2=-12x^2*y^2+3(a+6)^2-(a-6)^2=(a+6-(a-6))*(a+6+a-6)=(a+6-a+6)*2a=12*2a=24a(2x-3y)^2=4x^2-12xy+9y^2

№2:Хз сори(

№3

(a - 2b)² + 4b(a - b) = a² - 4ab + 4b² + 4ab - 4b² = a²

(a - 2b)² + 4b(a - b) = a² - 4ab + 4b² + 4ab - 4b² = a²Если a = 0,12 , то 0,12² = 0,0144

№4

1) 3(1+2xy)(1-2xy)=3(1-4x^ ^ 2y^ ^ 2)=3-12x^ ^ 2y^ ^ 2

1) 3(1+2xy)(1-2xy)=3(1-4x^ ^ 2y^ ^ 2)=3-12x^ ^ 2y^ ^ 23)(a+b)^ ^2-1 4ab a- b)^2=a^2+2ab+b^ -a^ 2+2ab-b^ ^ 2=

1) 3(1+2xy)(1-2xy)=3(1-4x^ ^ 2y^ ^ 2)=3-12x^ ^ 2y^ ^ 23)(a+b)^ ^2-1 4ab a- b)^2=a^2+2ab+b^ -a^ 2+2ab-b^ ^ 2=2)(x^ ^ 2-y^ ^ 3)^ ^ 2=x^ ^ 4-2x^ ^ 2y^ ^ 3+y^ ^ 6

№5

16y^ ^ 2=49

16y^ ^ 2=49ОТВЕТ: - 1.75n 1,

16y^ ^ 2=49ОТВЕТ: - 1.75n 1,y = - 7/4 * u y = 7/4

16y^ ^ 2=49ОТВЕТ: - 1.75n 1,y = - 7/4 * u y = 7/4y^ ^ 2=49/16