Пожалуйста помоги очень срочно надо)))

Пожалуйста помоги очень срочно надо)))
- Предмет:
  Алгебра
- Автор:
  janitor6lhs
- 5 лет назад

Ответы 2

Знаменатель не может быть равен нулю (на ноль делить нельзя)8. а) $\frac{5x-75x^2}{x^2 +8}; \ \ \ \ x^2 +8 eq 0; \ \ x^2 +8\ \textgreater \ 0 \\ \\ x \in (-\infty;+\infty)$ б) $\frac{5a-4}{a^2-100}; \ \ a^2 -100 eq 0; \ \ (a-10) \cdot (a+10) eq 0 \\ \\ a eq 10; \ \ a eq -10; \ \ x \in (-\infty; -10) \ \cup \ (-10; 10) \ \cup \ (10; +\infty)$ в) $\frac{y^4-81}{(y^3+6y^2) \cdot (y^2 -121)}; \ \ \ y^3 +6y^2 eq 0; \ \ y^2 \cdot (y+6) eq 0; \\ \\ y^2 eq 0; \ \ y eq 0; \ \ \ \ \ y eq -6; \\ (y^2-121) eq 0; \ \ (y-11) \cdot (y+11) eq 0; \ \ \ y eq 11; \ \ y eq -11 \\ \\ y \in (-\infty; -11) \ \cup \ (-11; -6) \ \cup \ (-6; 0) \ \cup \ (0;11) \ \cup \ (11; +\infty)$ г) $\frac{b+9}{b^2}; \ \ \ b^2 eq 0; \ \ b eq 0; \ \ \ b \in (-\infty;0) \ \cup \ (0; +\infty)$ д) $\frac{x-7x^5}{(3-7x) \cdot (2x+1)}; \ \ \ 3-7x eq 0; \ \ 7x eq 3; \ \ x eq \frac{3}{7}; \\\\ 2x+1 eq 0; \ \ 2x eq -1; \ \ \ x eq -\frac{1}{2} \\ \\ x \in (-\infty;-\frac{1}{2}) \ \cup \ (-\frac{1}{2}; \frac{3}{7}) \ \cup \ (\frac{3}{7}; + \infty)$ 9. а) $\frac{14a^4}{5xy} \cdot \frac{10x^2 y^3}{21 a^2 b^3}=\frac{2}{5 \cdot 3 b^3} \cdot a^{4-2} \cdot x^{2-1} \cdot y^{3-1}=\frac{2 \a^2xy^2}{15b^3}$ б) $\frac{25m^2 -4n^2 }{15mn} : \frac{25m^2 + 20mn + 4n^2}{9m^2}=\frac{(5m-2n) \cdot (5m+2n)}{15mn} \cdot \frac{9m^2}{(5m)^2 + 2 \cdot 5 \cdot 2 \cdot mn + (2n)^2}= \\ \\ =\frac{(5m-2n) \cdot (5m+2n)}{15mn} \cdot \frac{9m^2}{(5m+2n)^2 }=\frac{(5m-2n) \cdot 3m}{5n \cdot (5m+2n)}$ в) $\frac{3y^2}{x \cdot (x^3-y^3)}+\frac{y}{x \cdot (x^2 + xy + y^2) }-\frac{1}{x \cdot (x-y)}=\\ \\ = \frac{3y^2}{x \cdot (x-y) \cdot (x^2+xy+y^2 )}+\frac{y}{x \cdot (x^2 + xy + y^2) }-\frac{1}{x \cdot (x-y)}=\frac{3y^2 +y \cdot (x-y)-(x^2+xy+y^2)}{x \cdot (x-y) \cdot (x^2 +xy +y^2)} \\ \\ =\frac{3y^2 +xy -y^2 -x^2 -xy -y^2}{x \cdot (x-y) \cdot (x^2 +xy +y^2)}=\frac{y^2-x^2 }{x \cdot (x-y) \cdot (x^2 +xy +y^2)}=\frac{(y-x) \cdot (y+x)}{x \cdot (x-y) \cdot (x^2 +xy +y^2)}=$ $\\ \\ = -\frac{y+x}{x \cdot (x^2 +xy +y^2)}$ г) $\frac{(x+y) \cdot (x^2 -xy +y^2 )}{x-y} \cdot \frac{(x-y) \cdot (x^2 +xy+y^2)}{x+y}=(x^2-xy+y^2)\cdot(x^2+xy+y^2)=\\ \\ = x^4+x^2y^2 + y^4$ 10) а) $(\frac{8}{2a \cdot (a - 4)} -\frac{3a +32}{(a-4) \cdot (a^2 +4a+16)}) : \frac{a-8}{a \cdot (a^2 + 4a + 16)} -\frac{4}{-(a-4)}= \\ \\ =\frac{8\cdot (a^2 +4a+16)-2a \cdot (3a+32)}{2a \cdot (a-4) \cdot (a^2 +4a+16)} \cdot \frac{a \cdot (a^2 +4a+16)}{a-8}+\frac{4}{a-4}= \\ \\ = \frac{8a^2 +32a+128-6a^2 -64a}{2 \cdot (a-4)} \cdot \frac{1}{a-8}+\frac{4}{a-4} = \frac{2a^2 -32a+4 \cdot 2 \cdot(a-8)}{2 \cdot (a-4) \cdot (a-8) (a-4)}= \frac{2a^2 -32a+8a-64}{2 (a-4) (a-8)}\\ \\ = \frac{2 a^2 -24a-64}{2 (a-4) (a-8)}$ $\\ \\ = \frac{2 \cdo (a^2 -12a-32)}{2 (a-4) (a-8)}=\frac{(a-4) \cdot (a-8)}{(a-4) \cdot (a-8)}=1$ б) $\frac{(x+y) \cdot (x^2 -xy+y^2 )}{x+y} \cdot \frac{1}{(x-y) \cdot (x+y)} + \frac{2y}{x+y}-\frac{xy}{(x-y) \cdot (x+y)}=\\ \\ = \frac{x^2 -xy+y^2}{(x-y)(x+y)}+\frac{2y}{x+y}-\frac{xy}{(x-y) \cdot (x+y)}=\frac{x^2 -xy+y^2 +2y \cdot (x-y)-xy}{(x-y) \cdot (x+y)}= \\ \\ =\frac{x^2 -xy +y^2 +2xy-2y^2 -xy}{(x-y) \cdot (x+y)}=\frac{x^2 -y^2}{(x-y) \cdot (x+y)}=\frac{(x-y) \cdot (x+y)}{(x-y) \cdot (x+y)}=1$
- Автор:
  jaidynnote
- 5 лет назад
- 0
$8) \\ \\ 1) \frac{5x+75 x^{2} }{ x^{2} +8} \\ x^{2} +8 eq 0 \\ x^{2} eq -8 \\ x\in R \\ \\ 2) \frac{5a-4}{a ^{2} -100} \\ a ^{2} -100 eq 0 \\ a ^{2} eq 100 \\ a eq +-10 \\ a\in (- \infty ;-10) \cup (-10;10) \cup (10;+ \infty ) \\ \\$ $3) \frac{y ^{4}-81 }{(y ^{3}+6y ^{2} )(y ^{2} -121) } \\ (y ^{3}+6y ^{2} )(y ^{2} -121) eq 0 \\ y ^{3} +6y ^{2} eq 0 \\ y^{2} (y+6) eq 0 \\ y ^{2} eq 0 \\ y eq 0 \\ y+6 eq 0 \\ y eq -6 \\ y ^{2} -121=0 \\ y ^{2}=121 \\ y=+-11 \\ y\in (- \infty ;-11)\cup (-11;-6) \cup (-6;0) \cup (0;11)\cup \\ \cup(11;+\infty )$ $4) \frac{b+9}{b ^{2} } \\ b ^{2} eq 0 \\ b eq 0 \\ b\in (- \infty ;0)\cup (0;+\infty )$ $9. \\ \\ 1) \frac{14a ^{4}*10 x^{2} y^{3} }{5xy*21a ^{2} b ^{3} } = \frac{2a ^{2} *2xy ^{2} }{3b ^{3} } = \frac{4a ^{2} xy ^{2} }{3b ^{3} } \\ \\ 2) \frac{25m ^{2} -4n ^{2} }{15mn} * \frac{9m^2}{25m^2+20mn+4n^2} = \\ \\ = \frac{(5m-2n)(5m+2n)*9m^2}{15mn*(5m+2n)^2} = \frac{3m(5m-2n)}{5n(5m+2n)}$ $3) \frac{3y ^{2} }{ x^{4} -xy ^{3} } + \frac{y}{ x^{3}+ x^{2} y+xy ^{2} } - \frac{1}{ x^{2} -xy} = \frac{3y ^{2} }{x*(x^3-y^3)} + \frac{y}{x( x^{2} +xy+y^2)} - \\ \\ - \frac{1}{x(x-y)} = \frac{3y^2}{x(x-y)( x^{2} +xy+y^2)} + \frac{y}{x( x^{2} +xy+y^2)} - \frac{1}{x(x-y)} = \\ \\ = \frac{3y^2+y(x-y)-1*(x^2+xy+y^2)}{x(x-y)(x^2+xy+y^2)} = \frac{3y^2+xy-y^2-x^2-xy-y^2}{x(x^3-y^3)} = \\ \\ = \frac{y^2+xy-x^2-xy}{x(x^3-y^3)} = \frac{y(y+x)-x(x+y)}{x(x-y)(x^2+xy+y^2)} =$ $= \frac{(y+x)(y-x)}{x(x-y)(x^2+xy+y^2)} = \frac{-(x+y)(x-y)}{x(x-y)(x^2+xy+y^2)} =- \frac{x+y}{x(x^2+xy+y^2)}$ $4) \frac{x^3+y^3}{x-y}* \frac{x^3-y^3}{x+y} = \frac{(x+y)(x^2+xy+y^2)*(x-y)(x^-xy+y^2)}{(x-y)(x+y)} = \\ \\ =(x^2+xy-y^2)(x^2-xy+y^2)=x^4-x^3y+x^2y^2+\\ +x^3y-x^2y^2+xy^2+x^2y^2-xy^2+y^4=x^4+x^2y^2+y^4$ $10. \\ \\ 1)( \frac{8}{2a^2-8a} - \frac{3a+32}{a^3-64} ): \frac{a-8}{a^3+4a^2+16a} - \frac{4}{4-a} =1 \\ \\ 1)\frac{8}{2a^2-8a} - \frac{3a+32}{a^3-64}= \frac{8}{2a(a-4)} - \frac{a+32}{(a-4)(a^2+4a+16)} = \\ \\ = \frac{4(a^2+4a+16)-(3a+32)*a}{a(-4)(a^2+4a+16)} = \frac{4a^2+16a+64-3a^2-32a}{a(a-4)(a^2+4a+16)} = \\ \\ = \frac{a^2-16a+64}{a(a-4)(a^2+4a+16)} \\ \\$ $2) \frac{a^2-16a+64}{a(a-4)(a^2+4a+16)} * \frac{a(a^2+4a+16)}{(a-8)} = \frac{(a-8)^2*a(a^2+4a+16)}{a(a-4)(a^2+4a+16)(a-8)} = \\ \\ = \frac{a-8}{a-4} \\ \\ 3) \frac{a-8}{a-4} - \frac{4}{4-a} = \frac{a-8}{a-4} + \frac{4}{a-4} = \frac{a-8+4}{a-4} = \frac{a-4}{a-4} =1$ $2) \frac{x^3+y^3}{x+y} :(x^2-y^2)+ \frac{2y}{x+y} - \frac{xy}{x^2-y^2} =1 \\ \\ 1) \frac{x^3+y^3}{x+y} :(x^2-y^2)= \frac{(x+y)(x^2+xy+y^2)}{(x+y)(x+y)(x-y)} = \frac{ x^{2} +xy+y^2}{(x+y)(x-y)} \\ \\ 2) \frac{2y}{x+y} - \frac{xy}{( x^{2} -y ^{2} )} = \frac{2y}{x+y} - \frac{xy}{(x-y)(x+y)} = \frac{2y(x-y)-xy}{(x-y)(x+y)} = \\ \\ =\frac{2xy-2y^2-xy}{(x-y)(x+y)} \\ \\ 3) \frac{ x^{2} +xy+y^2+2xy-2y^2-xy}{(x-y)(x+y)} = \frac{ x^{2} -y^2}{ x^{2} -y^2} =1$
- Автор:
  bradyburns
- 5 лет назад
- 0
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