Question

темь зачет задачи ...........................​

Answer 1

$(x+5)^{2} -x*(x-10)=x^{2} +10x+25-x^{2} +10x=20x+25=20*(-\frac{1}{20} )+25=-1+25=24$

$(50b+9)(50b-9)-(50b+9)^{2} =(50b+9)(50b-9-(50b+9))=(50b+9)(50b-9-50b-9)=(50b+9)*(-18)=-18*(50b+9)=-18*(50*0,2+9)=-18*90=-1620$

$\frac{a{2}-ab }{2} +\frac{2ab-2a^{2}+b }{4} =\frac{2(a^{2}-ab)+2ab-2a^{2} +b }{4} =\frac{2a^{2}-2ab+2ab-2a^{2}+b }{4} =\frac{b}{4} =\frac{4}{4} =1$

$(\frac{1}{2a} +\frac{1}{6a})*\frac{a^{2} }{5} =\frac{4}{6a} *\frac{a^{2} }{5} =\frac{2}{3} *\frac{a}{5} =\frac{2a}{15}=\frac{2*(-4,8)}{15} =\frac{-9,6}{15} =-\frac{16}{25} =-0,64$

$\frac{42}{7a-a^{2} } -\frac{6}{a} =\frac{42-6(7-a)}{2*(7-a)} =\frac{42-42+6a}{a*(7-a)} =\frac{6a}{a*(7-a)} =\frac{6}{7-a } =\frac{6}{7-2} =\frac{6}{5} =1\frac{1}{5} =0,2$

$\frac{9b}{a-b} *\frac{a^{2}-ab }{18b} =\frac{1}{a-b} *\frac{a*(a-b)}{2}= \frac{a}{2} =\frac{81}{2} =40\frac{1}{2}=40,5$

$\frac{c^{2}-2ac }{c^{2} } :\frac{c-2a}{a}=\frac{c*(c-2a)}{a^{2} } *\frac{a}{c-2a}=\frac{c}{a}=\frac{46}{4}=11\frac{1}{2}=11,5$

$9b+\frac{5a-9b^{2} }{b}=\frac{9b^{2} +5a-9b^{2} }{b} =\frac{5a}{b}-\frac{5*9}{18}=\frac{45}{18} =2\frac{1}{2}=2,5$

$\frac{a^{2}-81 }{2a^{2} +18a} =\frac{(a-9)*(a+9)}{2a*(a+9)} =\frac{a-9}{2a} -\frac{-4,5-9}{2*(-4,5)} =\frac{-13,5}{-9} =1\frac{1}{2}=1,5$

$\frac{a^{2}-16b^{2} }{4ab} : (\frac{1}{4b} -\frac{1}{a} )=\frac{(a-4b)*(a+4b)}{4ab} :\frac{a-4b}{4ab} =\frac{(a-4b)*(a+4b)}{4ab}*\frac{4ab}{a-4b} =a+4b=3\frac{1}{13} +4*4\frac{3}{13} =3\frac{1}{13} +16\frac{12}{13} =20$