Упр.44 ГДЗ Муравин 7 класс (Алгебра)

1. Сравните значения выражений: 1) $$\left(x+\frac{5}{6}\right)\left(x-\frac{5}{6}\right)$$ и $$x+\frac{5}{6}\cdot\left(x-\frac{5}{6}\right)$$ при $$x=\frac{4}{9}$$; 2) $$-0{,}4(y+0{,}4)$$ и $$(y-0{,}4)y+0{,}4$$ при $$y=0{,}5$$.

1) При $$x=\frac{4}{9}$$:

$$\left(x+\frac{5}{6}\right)\left(x-\frac{5}{6}\right)=x^2-\left(\frac{5}{6}\right)^2$$
$$=\left(\frac{4}{9}\right)^2-\frac{25}{36} =\frac{16}{81}-\frac{25}{36} =\frac{64-225}{324} =-\frac{161}{324}$$

$$x+\frac{5}{6}\left(x-\frac{5}{6}\right) =\frac{4}{9}+\frac{5}{6}\left(\frac{4}{9}-\frac{5}{6}\right)$$
$$=\frac{4}{9}+\frac{5}{6}\left(\frac{8-15}{18}\right) =\frac{4}{9}+\frac{5}{6}\cdot\left(-\frac{7}{18}\right) =\frac{4}{9}-\frac{35}{108} =\frac{48}{108}-\frac{35}{108} =\frac{13}{108}$$

Так как $$-\frac{161}{324}<\frac{13}{108}$$, то

$$\left(x+\frac{5}{6}\right)\left(x-\frac{5}{6}\right)

2) При $$y=0{,}5$$:

$$-0{,}4(y+0{,}4)=-0{,}4\cdot(0{,}5+0{,}4)=-0{,}4\cdot0{,}9=-0{,}36$$

$$(y-0{,}4)y+0{,}4=(0{,}5-0{,}4)\cdot0{,}5+0{,}4=0{,}1\cdot0{,}5+0{,}4=0{,}05+0{,}4=0{,}45$$

Так как $$-0{,}36<0{,}45$$, то

$$-0{,}4(y+0{,}4)<(y-0{,}4)y+0{,}4.$$

Ответ

1) $$\left(x+\frac{5}{6}\right)\left(x-\frac{5}{6}\right)<x+\frac{5}{6}\left(x-\frac{5}{6}\right)$$; 2) $$-0{,}4(y+0{,}4)<(y-0{,}4)y+0{,}4$$.