Question

№53 помогите срочно нужно пж

Answer 1

Ответ:

Объяснение:

Я так понял, что нужно привести к общему знаменателю.

1) m/(2m+2n)=m/(2(m+n))

n/(8m-8n)=n/(8(m-n))=n/(2·4(m-n))

(mn)/(6m²-6n²)=(mn)/(6(m²-n²))=(mn)/(2·3(m-n)(m+n))

m/(2(m+n))=(m·4(m-n)·3)/(2(m+n)·4(m-n)·3)=(12m(m-n))/(24(m²-n²))

n/(8(m+n))=(n·3(m-n))/(8(m+n)·3(m-n))=(3n(m-n))/(24(m²-n²))

(mn)/(6(m²-n²))=(mn·4)/(6(m²-n²)·4)=(4mn)/(24(m²-n²))

2) (2c)/(5b-5c)=(2c)/(5(b-c))

(3a²)/(35b²-35c²)=(3a²)/(35(b²-c²))=(3a²)/(5·7(b-c)(b+c))

(7b)/(14b+14c)=(7b)/(14(b+c))=b/(2(b+c))

(2c)/(5(b-c))=(2c·7(b+c)·2)/(5(b-c)·7(b+c)·2)=(28c(b+c))/(70(b²-c²))

(3a²)/(35(b²-c²))=(3a²·2)/(35(b²-c²)·2)=(6a²)/(70(b²-c²))

b/(2(b+c))=(b·5(b-c)·7)/(2(b+c)·5(b-c)·7)=(5b(b-c))/(70(b²-c²))

3) 1/(a²-4b²)=1/((a-2b)(a+2b))

1/(3a²+6ab)=1/(3a(a+2b))

1/(2ab-a²)=1/(a(2b-a))=-1/(a(a-2b))

1/(a²-4b²)=(1·3a)/((a²-4b²)·3a)=(3a)/(3a(a²-4b²))

1/(3a(a+2b))=(1(a-2b))/(3a(a+2b)(a-2b))=(a-2b)/(3a(a²-4b²))

-1/(a(a-2b))=(-1·3(a+2b))/(a(a-2b)·3(a+2b))=(-3(a+2b))/(3a(a²-4b²))

4) 5/(4x-4)=5/(4(x-1))

(4x)/(1-x²)=(-4x)/(x²-1)=(-4x)/((x-1)(x+1))

1/(3x²+3x)=1/(3x(x+1))

5/(4(x-1))=(5(x+1)·3x)/(4(x-1)(x+1)·3x)=(15x(x+1))/(12x(x²-1))

(-4x)/(x²-1)=(-4x·4·3x)/((x²-1)·4·3x)=(-48x)/(12x(x²-1))

1/(3x(x+1))=(1·4(x-1))/(3x(x+1)·4(x-1)=(4(x-1))/(12x(x²-1))