Help meнайдите значение выражения х^2+1/х^2,если х+1/х=3; представьте в виде дроби выражение 1)х^-3+y^-2,2)c^-5-c^o,3)d^0-d^-6; 4)xy^-5-zy^-2; 5)a^-2b^-1+a^-1b^-2; 6)(a+a^-1)(b+b^-1); 7)(m^2+m^-1)(n^2+n^-2); 8)(x^-1-y^-1): (x^-2-y^-2).2.2.25возвести в куб: 1)(2а^2+b)^3; 2)(a-2b^2)^3; 3)(m^2/3-n^3)3; 4)m^2+n^2/3)3^; 5)(a^n+b^n)^3; 6)(a^n-2-b^n+2)^3

Ответ:

ZhannaMeer

08.07.2020 10:19

$\frac{x+1}{x}=3 \\ x+1=3x \\ 2x=1 \\ x= \frac{1}{2} \\ \\ 
 \frac{x^2+1}{x^2}= \frac{(\frac{1}{2})^2+1}{(\frac{1}{2})^2}=\frac{\frac{1}{4}+1}{\frac{1}{4}}= \frac{5}{4} : \frac{1}{4} =\frac{5}{4} * 4=5$

$1)x^{-3}+y^{-2}=\frac{1}{x^3}+\frac{1}{y^2}= \frac{y^2+x^3}{x^3y^2} \\ \\ 
2)c^{-5}-c^0=\frac{1}{c^5}-1=\frac{1-c^5}{c^5} \\ \\ 
3)d^0-d^{-6}=1-\frac{1}{d^6}=\frac{d^6-1}{d^6} \\ \\ 
4)xy^{-5}-zy^{-2}=\frac{x}{y^5}-\frac{z}{y^2}=\frac{xy^2-zy^5}{y^{10}}=\frac{y^2(x-zy^3)}{y^{10}}=\frac{x-zy^3}{y^5}$

$5)a^{-2}b^{-1}+a^{-1}b^{-2}= \frac{1}{a^2b}+\frac{1}{ab^2}=\frac{b+a}{a^2b^2} \\ \\ 
6)(a+a^{-1})(b+b^{-1})=(a+\frac{1}{a})(b+\frac{1}{b})= \\ \\ =\frac{a^2+1}{a}*\frac{b^2+1}{b}= \frac{(a^2+1)(b^2+1)}{ab}=\frac{a^2b^2+a^2+b^2+1}{ab} \\ \\ 
7)(m^2+m^{-1})(n^2+n^{-2})=(m^2+\frac{1}{m})(n^2+\frac{1}{n^2})= \\ \\ 
=\frac{m^3+1}{m}*\frac{n^4+1}{n^2}=\frac{(m^3+1)(n^4+1)}{mn^2}=\frac{m^3n^4+m^3+n^4+1}{mn^2}$

$8)(x^{-1}-y^{-1}):(x^{-2}-y^{-2})=(\frac{1}{x}-\frac{1}{y}):(\frac{1}{x^2}-\frac{1}{y^2})= \\ \\ 
=\frac{y-x}{xy}:\frac{y^2-x^2}{x^2y^2}=\frac{y-x}{xy}*\frac{x^2y^2}{y^2-x^2}=\frac{y-x}{xy}*\frac{x^2y^2}{(y-x)(y+x)}=\frac{xy}{y+x}$

$1)(2a^2+b)^3=(2a^2+b)(2a^2+b)(2a^2+b)= \\ =(4a^4+2a^2b+2a^2b+b^2)(2a^2+b)=(4a^4+4a^2b+b^2)(2a^2+b)= \\ =8a^6+4a^4b+8a^4b+4a^2b^2+2a^2b^2+b^3= \\ 
=8a^6+12a^4b+6a^2b^2+b^3 \\ \\ 
2)(a-2b^2)^3=(a-2b^2)(a-2b^2)(a-2b^2)= \\ 
=(a^2-2ab^2-2ab^2+4b^4)(a-2b^2)=(a^2-4ab^2+4b^4)(a-2b^2)= \\ 
=a^3-2a^2b^2-4a^2b^2+8ab^4+4ab^4-8b^6= \\ 
=a^3-6a^2b^2+12ab^4-8b^6$

$3)(\frac{m^2}{3}-n^3)^3=(\frac{m^2-3n^3}{3})^3=\frac{(m^2-3n^3)^3}{3^3}=\frac{(m^2-3n^3)(m^2-3n^3)(m^2-3n^3)}{27}= \\ \\ =\frac{(m^4-3m^2n^3-3m^2n^3+9n^6)(m^2-3n^3)}{27}=\frac{(m^4-6m^2n^3+9n^6)(m^2-3n^3)}{27}= \\ \\ =\frac{m^6-3m^4n^3-6m^4n^3+18m^2n^6+9m^2n^6-27n^9}{27}=\frac{m^6-9m^4n^3+27m^2n^6-27n^9}{27} \\ \\ 
$

$4)(m^2+\frac{n^2}{3})^3=(\frac{3m^2+n^2}{3})^3=\frac{(3m^2+n^2)^3}{3^3}=\frac{(3m^2+n^2)(3m^2+n^2)(3m^2+n^2)}{27}= \\ \\ =\frac{(9m^4+3m^2n^2+3m^2n^2+n^4)(3m^2+n^2)}{27}=\frac{(9m^4+6m^2n^2+n^4)(3m^2+n^2)}{27}= \\ \\ 
=\frac{27m^6+9m^4n^2+18m^4n^2+6m^2n^4+3m^2n^4+n^6}{27}=\frac{27m^6+27m^4n^2+9m^2n^4+n^6}{27}$

$5)(a^n+b^n)^3=(a^n+b^n)(a^n+b^n)(a^n+b^n)= \\ 
=(a^{2n}+a^nb^n+a^nb^n+b^{2n})(a^n+b^n)= \\ 
=(a^{2n}+2a^nb^n+b^{2n})(a^n+b^n)= \\
=a^{3n}+a^{2n}b^n+2a^{2n}b^n+2a^nb^{2n}+a^nb^{2n}+b^{3n}= \\ 
=a^{3n}+3a^{2n}b^n+3a^nb^{2n}+b^{3n} \\ \\ 
$

$6)(a^{n-2}-b^{n+2})^3=(a^n*a^{-2}-b^n*b^2)^3=(\frac{a^n}{a^2}-b^n*b^2)^3= \\ \\ =(\frac{a^n-a^2*b^n*b^2}{a^2})^3=\frac{a^n-a^2b^2b^n}{a^2}*\frac{a^n-a^2b^2b^n}{a^2}*\frac{a^n-a^2b^2b^n}{a^2}= \\ \\ =\frac{(a^n-a^2b^2b^n)(a^n-a^2b^2b^n)}{a^4}*\frac{a^n-a^2b^2b^n}{a^2}= \\ \\ =\frac{a^{2n}-2a^2b^2a^nb^n+a^4b^4b^{2n}}{a^4}*\frac{a^n-a^2b^2b^n}{a^2}= \\ \\ =\frac{a^{3n}-3a^2b^2a^{2n}b^n+3a^4b^4a^nb^{2n}-a^6b^6b^{3n}}{a^6}=$

$=\frac{a^{3n}-3a^{2n+2}b^{n+2}+3a^{n+4}b^{2n+4}-a^6b^{3n+6}}{a^6}= \\ \\ 
=a^{3n-6}-3a^{2n-4}b^{n+2}+3a^{n-2}b^{2n+4}-b^{3n+6}$