Question

Помоги плиз. Если можно, то всё.

Answer 1

$1) \frac{6}{x^2+3x}- \frac{2}{x}= \frac{6-2x-6}{x(x+3)}=- \frac{2x}{x(x+3)}=- \frac{2}{x+3}$
$\frac{5b}{a^2-ab}* \frac{a^2-b^2}{10b^2}= \frac{(a-b)(a+b)}{a(a-b)*2} = \frac{a+b}{2a}$
$2) \frac{x+1}{5}+ \frac{x-1}{4}=1 \\ \frac{4(x+1)+5(x-1)-20}{20}=0 \\ 4x+4+5x-5-20=0 \\ 9x=21 \\ x= \frac{21}{9}=2 \frac{1}{3}$
$3) (2.3+10^9)(3*10^{-12})= \frac{2.3+10^9*3}{10^{12}} = \frac{6.9*10^9}{10^{12}} = \frac{6.9}{10^3}=0.0069 \ \textgreater \ 0.006$
$4) 3 \sqrt{2}* \frac{1}{2} \sqrt{8}= \sqrt{18} * \sqrt{ \frac{8}{4} } = \sqrt{ \frac{18*8}{4} } = \sqrt{36}=6$
$5) \frac{16^{-2}*27^{-4}}{6^{-12}}= \frac{6^{12}}{16^2*27^4}= \frac{2^{12}*3^{12}}{2^2*2^2*2^2*2^2*3^4*3^4*3^4}=2^4=16$
$6) \sqrt{45-20 \sqrt{5} }= \sqrt{25+20-20 \sqrt{5} }= \sqrt{5^2-20 \sqrt{5}+20 } = \\ \sqrt{(5-2 \sqrt{5})^2} =5-2 \sqrt{5}$
др.листик
$3) \frac{a^3*a^{-12}}{a^{-6}}= \frac{a^3*a^6}{a^{12}} = \frac{1}{a^3} \\ 1:( \frac{1}{2})^3=1*2^3=8$
$4) 2 \sqrt{3}\ \textgreater \ \sqrt{11} \\ \sqrt{12}\ \textgreater \ \sqrt{11}$
$5) 23*10^{-5}= \frac{23}{10^5} \\ 2.7*10^{-6}= \frac{2.7}{10^6}= \frac{0.27}{10^5} \\ 210*10^{-6}= \frac{210}{10^6}= \frac{21}{10^5} \\ 2.7*10^{-6}; 210*10^{-6}; 23*10^{-5}$