Решение №5806: \(\frac{(x+1)^{3}}{x}-\frac{(x+1)^{2}}{x+2}-\frac{1}{x}+\frac{1}{x+2}=\frac{(x+1)^{3}}{x}-\frac{1}{x}-\frac{(x+1)^{2}}{x+2}+\frac{1}{x+2}=\frac{(x+1)^{3}-1}{x}=-(\frac{(x+1)^{2}-1}{x+2})=\frac{x^{3}+3x^{2}+3x+1-1}{x}-\frac{x^{2}=2x+1-1}{x+2}=\frac{x(x^{2}+3x+3)}{x}-\frac{x^{2}+2x}{x+2}=x^{2}+3x+3-\frac{x(x+2)}{x+2}=x^{2}+3x+3-x=x^{2}+2x+3=(x^{2}+2x+1)+2=(x+1)^{2}+2; x=-2,1; (-2,1+1)^{2}+2=(-1,1)^{2}+2=1,21+2=3,21\)

Ответ: \(3,21\)

Решение №5807: \( f(x)-f(-x)=\frac{2x+3}{x^{2}-x+1}+\frac{3-2x}{x^{2}+x+1}-(\frac{2x+3}{x^{2}-x+1}+\frac{3-2x}{x^{2}+x+1})=\frac{2x+3}{x^{2}-x+1}+\frac{3-2x}{x^{2}+x+1}-\frac{3-2x}{x^{2}+x+1}-\frac{3+2x}{x^{2}-x+1}=\frac{2x+3}{x^{2}-x+1}-\frac{3+2x}{x^{2}-x+1}+\frac{3-2x}{x^{2}+x+1}-\frac{3-2x}{x^{2}-x+1}=\frac{2x+3}{x^{2}-x+1}-\frac{3+2x}{x^{2}-x+1}+\frac{3-2x}{x^{2}+x+1}-\frac{3-2x}{x^{2}+x+1}-\frac{3-2x}{x^{2}+x+1}=\frac{2x+3-3-2x}{x^{2}-x+1}+\frac{3-2x-3+2x}{z^{2}+x+1}=0+0=0\)

Ответ: 0

Решение №5808: \(f(x)+f(-x)=\frac{2x^{2}+3x}{3x^{2}-7x+5}-\frac{2x^{2}-3x}{3x^{2}+7x+5}+(\frac{2x^{2}+3x}{3x^{2}-7x+5}-\frac{2x^{2}-3x}{3x^{2}+7x+5})=\frac{2x^{2}+3x}{3x^{2}-7x+5}-\frac{2x^{2}-3x}{3x^{2}+7x+5}+(\frac{2x^2}-3x}{3x^{2}+7x+5}-\frac{2n^{2}+3x}{3x^{2}-7x+5})=\frac{2x^{2}+3x}{3x^{2}-7x+5}-\frac{2x^{2}+3x}{3x^{2}-7x+5}-\frac{2x^2}-3x}{3x^{2}+7x+5}-\frac{2n^{2}-3x}{3n^{2}+7x+5}=\frac{2x^{2}+3x-2x^{2}-3x}{3x^{2}-7x+5}+\frac{2x^{2}-3x-2x^{2}+3x}{3x^{2}+7x+5}=0+0=0\)

Ответ: 0

Решение №5809: \(x=\frac{a-b}{a+b}; y=\frac{b-c}{b+c}; z=\frac{c-a}{c+a}; (1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)=(1+\frac{a-b}{a+b})(1+\frac{b-c}{b+c})(1+\frac{c-a}{c+a})=(\frac{a+b+a-b}{a+b})(\frac{b+c+b-c}{b+c})(\frac{c+a+c-a}{c+a})=\frac{2a}{a+b} \cdot \frac{2b}{b+c} \cdot \frac{2c}{c+a} =\frac{8abc}{(a+b)(b+c)(c+a)}; (1-x)(1-y)(1-z)=(1-\frac{a-b}{a+b})(1-\frac{b-c}{b+c})(1-\frac{c-a}{c+a})=(\frac{a+b-a+b}{a+b})(\frac{b+c-b+c}{b+c})(\frac{c+a-c+a}{c+a})=\frac{2b}{a+b} \cdot \frac{2c}{b+c} \cdot \frac{2a}{c+a} =\frac{8abc}{(a+b)(b+c)(c+a)}; \frac{8abc}{(a+b)(b+c)(c+a)}=\frac{8abc}{(a+b)(b+c)(c+a)}, значит (1+z)(1+y)(1+z)=(1-x)(1-y)(1-y)(1-z)\)

Ответ: \((1-x)(1-y)(1-y)(1-z)\)

Решение №5810: \(\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-a)(b-c)}+\frac{c^{3}}{(c-a)(c-b)}=\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-a)(b-c)}+\frac{c^{3}}{(c-a)(c-b)}=\frac{a^{3}}{(a-b)(a-c)}-\frac{b^{3}}{(a-b)(b-c)}+\frac{c^{3}}{(a-b)(b-c)}=\frac{a^{2}(b-c)-b^{3}(a-c)+c^{3}(a-b)}{(a-b)(a-c)(b-c)}=\frac{a^{3}b-a^{3}c-ab^{3}+b^{3}+b^{3}c+ac^{3}-bc^{3}}{(a-b)(a-c)(b-c)}=\frac{ab(a^{2}-b^{2})+c(b^{3}-a^{3})+c^{3}(a-b)}{(a-b)(a-c)(b-c)}=\frac{ab(a-b)(a+b)-c(a-b)(a^{2}+ab+b^{2})+c^{3}(a-b)}{(a-b)(a-c)(b-c)}=\frac{ab(a-b)(a+b)-c(a-b)(a^{2}+ab+b^{2})+c^{3}(a-b)}{(a-b)(a-c)(b-c)}=\frac{(a-b)(ab(a+b)-c(a^{2}+ab+b^{2})+c^{3})}{(a-b)(a-c)(b-c)}=\frac{a^{2}b+ab^{2}-a^{2}c-abc-b^{2}c+c^{3}}{(a-c)(b-c)}=\frac{a^{2}(b-c)+ab(b-c)-c(b^{2}-c^{2})}{(a-c)(b-c)}=\frac{a^{2}(b-c)+ab(b-c)-c(b-c)(b+c)}{(a-c)(b-c)}=\frac{(b-c)(a^{2}+ab-c(b+c))}{(a-c)(b-c)}=\frac{a^{2}+ab-cb-c^{2}}{a-c}=\frac{a^{2}-c^{2}+ab-bc}{a-c}=\frac{(a-c)(a+c)+b(a-c)}{a-c}=\frac{(a-c)(a+c)+b(a-c)}{a-c}=\frac{(a-c)(a+c+b)}{a-c}=a+b+c\)

Ответ: \(a+b+c\)

Решение №5812: \(\frac{b-c}{b+c}+\frac{c-a}{c+a}+\frac{a-b}{a+b}+\frac{(b-c)(c-a)(a-b)}{(b+c)(c+a)(a+b)}=\frac{(b-c)(c+a)(a+b)+(c-a)(b+c)(a+b)-(a-b)(b+c)(c+a)-(b-c)(c-a)(a-b)}{(b+c)(c+a)(a+b)}=\frac{(b-c)(ac+bc+a^{2}+ab+ac-bc-a^{2}+ab)+(b+c)(ac+bc-a^{2}-ab+a^{2}+ac-bc-ab)}{(b+c)(c+a)(a+b)}=\frac{(b-c)(2ac+2ab)+(b+c)(2ac-2ab)}{(b+c)(c+a)(a+b)}=\frac{2abc+2ab^{2}-2ac^{2}-2abc+2abc-2ab^{2}+2ac^{2}-2abc}{(b+c)(c+a)(a+b)}=\frac{0}{((b+c)(c+a)(a+b)}=0\)

Ответ: 0

Решение №5813: \(\frac{1}{n}-\frac{1}{n+1}=\frac{1 \cdot (n+1)-1 \cdot n}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{1}{n(n+1)}; \frac{1}{n(n+1)}=\frac{1}{n(n+1)}\)

Ответ: NaN

Решение №5815: \(\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{1}{2} \cdot \frac{2n+1-2n+1}{(2n-1)(2n+1)}=\frac{1}{2} \cdot \frac{2}{(2n-1)(2n+1)}=\frac{1}{(2n-1)(2n+1)}\)

Ответ: NaN

Решение №5817: \(\frac{6a}{b}:\frac{3a}{b}=frac{6a}{b} \cdot \frac{b}{3a}=\frac{6a \cdot b}{b \cdot 3a}=2\)

Ответ: \(2\)

Решение №5818: \(-\frac{4p}{q} \cdot \frac{q}{2p}=-frac{4p \cdot q}{q \cdot 2p}=-2\)

Ответ: \(-2\)