Упр.341 ГДЗ Никольский Потапов 7 класс (Алгебра)
а) (1/2 + a)2;
б) (x+1/3)2;
в) (m+0,2)2;
г) (1,1+p)2;
д) (1/2*a+2/3*b)2;
е) (3/4*x+1/5*y)2;
ж) (0,2m+2,1n)2;
з) (0,4p+9,3q)2;
и) (3/5*ab + 1/2*c2)2.
Используем формулу квадрата суммы:
$$ (u+v)^2=u^2+2uv+v^2. $$
$$\left(\frac12+a\right)^2=\left(\frac12\right)^2+2\cdot\frac12\cdot a+a^2=\frac14+a+a^2.$$
$$\left(x+\frac13\right)^2=x^2+2\cdot x\cdot\frac13+\left(\frac13\right)^2=x^2+\frac23x+\frac19.$$
$$\left(m+0,2\right)^2=m^2+2\cdot m\cdot0,2+0,2^2=m^2+0,4m+0,04.$$
$$\left(1,1+p\right)^2=1,1^2+2\cdot1,1\cdot p+p^2=1,21+2,2p+p^2.$$
$$\left(\frac12 a+\frac23 b\right)^2=\left(\frac12 a\right)^2+2\cdot\frac12 a\cdot\frac23 b+\left(\frac23 b\right)^2=\frac14a^2+\frac23ab+\frac49b^2.$$
$$\left(\frac34 x+\frac15 y\right)^2=\left(\frac34 x\right)^2+2\cdot\frac34 x\cdot\frac15 y+\left(\frac15 y\right)^2=\frac{9}{16}x^2+\frac{3}{10}xy+\frac{1}{25}y^2.$$
$$\left(0,2m+2,1n\right)^2=(0,2m)^2+2\cdot0,2m\cdot2,1n+(2,1n)^2=0,04m^2+0,84mn+4,41n^2.$$
$$\left(0,4p+0,3q\right)^2=(0,4p)^2+2\cdot0,4p\cdot0,3q+(0,3q)^2=0,16p^2+0,24pq+0,09q^2.$$
$$\left(\frac35ab+\frac12c^2\right)^2=\left(\frac35ab\right)^2+2\cdot\frac35ab\cdot\frac12c^2+\left(\frac12c^2\right)^2=\frac{9}{25}a^2b^2+\frac35abc^2+\frac14c^4.$$
Ответ
а) $$\frac14+a+a^2$$;
б) $$x^2+\frac23x+\frac19$$;
в) $$m^2+0,4m+0,04$$;
г) $$1,21+2,2p+p^2$$;
д) $$\frac14a^2+\frac23ab+\frac49b^2$$;
е) $$\frac{9}{16}x^2+\frac{3}{10}xy+\frac{1}{25}y^2$$;
ж) $$0,04m^2+0,84mn+4,41n^2$$;
з) $$0,16p^2+0,24pq+0,09q^2$$;
и) $$\frac{9}{25}a^2b^2+\frac35abc^2+\frac14c^4$$.
