Помогите с номером 341, пожалуйста.

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Дано ответов: 2

АннаАрт_zn · Answer 1 · 2018-03-18T20:02:43+0000

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

$b) (x+ \frac{1}{3} )^2=(x+ \frac{1}{3} )(x+ \frac{1}{3} )=(x)(x)+(x)(\frac{1}{3})+(\frac{1}{3})(x)+(\frac{1}{3})(\frac{1}{3})= \\ \\ =x^2+\frac{1}{3}x+\frac{1}{3}x+\frac{1}{9}=x^2+\frac{2}{3}x+\frac{1}{9}$

$v) (m+0,2)^2=(m+ \frac{2}{10} )^2=(m+ \frac{2}{10} )(m+ \frac{2}{10} )= \\ \\ =(m)(m)+(m)(\frac{2}{10})+(\frac{2}{10})(m)+(\frac{2}{10})(\frac{2}{10})= \\ \\ =m^2+\frac{2}{10}m+\frac{2}{10}m+ \frac{4}{100} =m^2+\frac{4}{10}m+ \frac{4}{100}= \\ \\ =m^2+0,4m+0,04$

$g) (1,1+p)^2=(1 \frac{1}{10} +p)^2=( \frac{11}{10} +p)^2=( \frac{11}{10} +p)( \frac{11}{10} +p)= \\ \\ =(\frac{11}{10})(\frac{11}{10})+(\frac{11}{10})(p)+(p)(\frac{11}{10})+(p)(p)= \\ \\ = \frac{121}{100} +\frac{11}{10}p+\frac{11}{10}p+p^2=\frac{121}{100} +\frac{22}{10}p+p^2= \\ \\ =1,21 +2,2p+p^2$

$d) ( \frac{1}{2}a+ \frac{2}{3}b )^2=( \frac{1}{2}a+ \frac{2}{3}b )( \frac{1}{2}a+ \frac{2}{3}b )= \\ \\ =(\frac{1}{2}a)(\frac{1}{2}a)+(\frac{1}{2}a)(\frac{2}{3}b)+(\frac{2}{3}b)(\frac{1}{2}a)+(\frac{2}{3}b)(\frac{2}{3}b)= \\ \\ = \frac{1}{4} a^2+ \frac{1}{3} ab+ \frac{1}{3} ab+ \frac{4}{9} b^2= \frac{1}{4} a^2+ \frac{2}{3} ab+ \frac{4}{9} b^2$

$e) ( \frac{3}{4} x+ \frac{1}{5} y)^2=( \frac{3}{4} x+ \frac{1}{5} y)( \frac{3}{4} x+ \frac{1}{5} y)= \\ \\ =(\frac{3}{4} x)(\frac{3}{4} x)+(\frac{3}{4} x)(\frac{1}{5} y)+(\frac{1}{5} y)(\frac{3}{4} x)+(\frac{1}{5} y)(\frac{1}{5} y)= \\ \\ = \frac{9}{16} x^2+ \frac{3}{20} xy+ \frac{3}{20} xy+ \frac{1}{25} y^2= \frac{9}{16} x^2+\frac{3}{10} xy+ \frac{1}{25} y^2$

$gg) (0,2m+2,1n)^2=( \frac{2}{10} m+2 \frac{1}{10} n)^2=( \frac{2}{10} m+ \frac{21}{10} n)^2= \\ \\ =( \frac{2}{10} m+ \frac{21}{10} n)( \frac{2}{10} m+ \frac{21}{10} n)= \\ \\ =(\frac{2}{10} m)(\frac{2}{10} m)+(\frac{2}{10} m)(\frac{21}{10} n)+(\frac{21}{10} n)(\frac{2}{10} m)+(\frac{21}{10} n)(\frac{21}{10} n)= \\ \\ = \frac{4}{100} m^2+ \frac{42}{100} mn+\frac{42}{100} mn+ \frac{441}{100} n^2=\frac{4}{100} m^2+\frac{84}{100} mn+ \frac{441}{100} n^2=$

$=0,04m^2+0,84 mn+ 4,41 n^2$

$z) (0,4p+0,3q)^2=( \frac{4}{10} p+ \frac{3}{10} q)^2=( \frac{4}{10} p+ \frac{3}{10} q)( \frac{4}{10} p+ \frac{3}{10} q)= \\ \\ =(\frac{4}{10} p)(\frac{4}{10} p)+(\frac{4}{10} p)(\frac{3}{10} q)+(\frac{3}{10} q)(\frac{4}{10} p)+(\frac{3}{10} q)(\frac{3}{10} q)= \\ \\ = \frac{16}{100} p^2+ \frac{12}{100} pq+\frac{12}{100} pq+ \frac{9}{100} q^2=\frac{16}{100} p^2+\frac{24}{100} pq+ \frac{9}{100} q^2= \\ \\ =0,16p^2+0,24pq+0,09q^2$

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