Question

boberon, докажите тождество:
1. (b³-1/b-1 + b) : b²-1/b-1 = b+1;
2. 1+b/1-b² * (1+b³/1+b - b) = 1-b.

сократите дробь:
a⁴-4/a³+2a
x⁴-4x²+4/x³-2x
x⁴-6x²+9/3x-x³
x²-2x+4/x³+8

Answer 1

1. ((b³-1)/(b-1) + b) :( b²-1)/(b-1 )= ((b-1)*(b²+b+1)/(b-1) + b) :(( b-1)(b+1)/(b-1 )=

(b²+b+1+b) :(b+1)=(b+1)² :(b+1)=b+1;  доказано.

2. (1+b)/(1-b² )* ((1+b³)/(1+b) - b) =((1+b)/((1-b)*(1+b)))* ((1+b)*(1-b+b²)/(1+b) - b) =

(1/(1-b)*)* ((1-b+b²- b) =1*(1-b)²/(1-b)= 1-b. доказано.

 

сократите дробь:

(a⁴-4)/(a³+2a)=(a²+2)(a²-2)/(a(a²+2))=(a²-2)/a

(x⁴-4x²+4)/(x³-2x)=(x²-2)²/(x*(x²-2))=(x²-2)/x

(x⁴-6x²+9)/(3x-x³)=(3-x²)²/(x*(3-x²))=(3-x²)/x

(x²-2x+4)/(x³+8)=(x²-2x+4)/((x+2)*(x²-2x+4))=1/(x+2)

Answer 2

$\left( \dfrac{b^3-1}{b-1} + b \right) : \dfrac{b^2-1}{b-1} = \left( \dfrac{(b-1)(b^2+b+1)}{b-1} + b \right) : \dfrac{(b-1)(b+1)}{b-1} = (b^2 + b + 1 + b) : (b + 1) = (b+1)^2 : (b+1) = b + 1$

$\dfrac{1+b}{1-b^2} \left(\dfrac{1+b^3}{1+b} - b\right) = \dfrac{1+b}{(1-b)(1+b)}\left(\dfrac{(1+b)(1-b+b^2)}{1+b} - b\right) = \dfrac{1}{1-b} (1-b+b^2-b) = \dfrac{(1-b)^2}{1-b} = 1-b$

$\dfrac{a^4-4}{a^3+2a} = \dfrac{(a^2-2)(a^2+2)}{a(a^2+2)}= \dfrac{a^2-2}{a}$

$\dfrac{x^4-4x^2+4}{x^3-2x} = \dfrac{(x^2 - 2)^2}{x(x^2-2)}=\dfrac{x^2-2}{x}$

$\dfrac{x^4-6x^2+9}{3x-x^3} = \dfrac{(x^2-3)^2}{x(3-x^2)} = \dfrac{(3-x^2)^2}{x(3-x^2)} = \dfrac{3-x^2}{x}$

$\dfrac{x^2-2x+4}{x^3+8} = \dfrac{x^2-2x+4}{(x+2)(x^2-2x+4)} = \dfrac{1}{x+2}$