Question

Помогите с номером 341, пожалуйста.

Answer 1

$1) ( frac{1}{2} +a)^{2} = frac{1}{4} +a ^{2} +a \ 2)(x+ frac{1}{3} )^{2} = x^{2} + frac{1}{9} + frac{2}{3} x \ 3)(m+0.2) ^{2} =m^{2} +0.04+0.4m \ 4)(1.1+p) ^{2} =1.21+p ^{2} +2.2p \ 5)( frac{1}{2} a+ frac{2}{3} b)^{2} = frac{1}{4} a^{2} + frac{4}{9} b^{2} + frac{2}{3} ab \ 6)( frac{3}{4} x+ frac{1}{5} y)^{2} =(0.75x+0.2y)^{2} =0.5625 x^{2} +0.04y ^{2} +0.3xy \ 7)(0.2m+2.1n)^{2} =0.04m^{2} +4.41n ^{2} +0.84mn \ 8)(0.4p+0.3q) ^{2}=0.16p^{2} +0.09q^{2}+0.24pq$
$9) ( frac{3}{5} ab+ frac{1}{2} c^{2} )^{2} =(0.6ab+0.5c^{2} )^{2} = 0.36a ^{2} b ^{2} +0.25c ^{4} +0.6abc^{2}$

Answer 2

А откуда во 2 примере появилось 2/3х?

Answer 3

Тогда уже формула по другому пишется.
(а+в)^2=a^2+2ab+b^2

Answer 4

от перестановки мест слагаемых сумма не меняется

Answer 5

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

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

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

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

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

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

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

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

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

$i) ( frac{3}{5} ab+ frac{1}{2} c^2)^2=( frac{3}{5} ab+ frac{1}{2} c^2)( frac{3}{5} ab+ frac{1}{2} c^2)= \ \ =(frac{3}{5} ab)(frac{3}{5} ab)+(frac{3}{5} ab)(frac{1}{2} c^2)+(frac{1}{2} c^2)(frac{3}{5} ab)+(frac{1}{2} c^2)(frac{1}{2} c^2)= \ \ = frac{9}{25} a^2b^2+ frac{3}{10} abc^2+ frac{3}{10} abc^2+ frac{1}{4} c^4= frac{9}{25} a^2b^2+frac{3}{5} abc^2+ frac{1}{4} c^4$