Упр.899 ГДЗ Макарычев 7 класс (Алгебра)

а) x2 — y2;
б) с2- z2;
в) а2 — 25;
г) m2 — 1;
д) 16 — b2;
е) 100 — x2;
ж) р2 — 400;
з) y2 — 0,09;
и) 1,44 -a2;
К) b2- 4/9;
д) 9/16 — n2;
м) 25/49 — p2.

Используем формулу разности квадратов:

$$a^2-b^2=(a-b)(a+b).$$

1. $$x^2-y^2=(x-y)(x+y)$$
2. $$c^2-z^2=(c-z)(c+z)$$
3. $$a^2-25=a^2-5^2=(a-5)(a+5)$$
4. $$m^2-1=(m-1)(m+1)$$
5. $$16-b^2=4^2-b^2=(4-b)(4+b)$$
6. $$100-x^2=10^2-x^2=(10-x)(10+x)$$
7. $$p^2-400=p^2-20^2=(p-20)(p+20)$$
8. $$y^2-0{,}09=y^2-0{,}3^2=(y-0{,}3)(y+0{,}3)$$
9. $$1{,}44-a^2=1{,}2^2-a^2=(1{,}2-a)(1{,}2+a)$$
10. $$b^2-\frac{4}{9}=b^2-\left(\frac{2}{3}\right)^2=\left(b-\frac{2}{3}\right)\left(b+\frac{2}{3}\right)$$
11. $$\frac{9}{16}-n^2=\left(\frac{3}{4}\right)^2-n^2=\left(\frac{3}{4}-n\right)\left(\frac{3}{4}+n\right)$$
12. $$\frac{25}{49}-p^2=\left(\frac{5}{7}\right)^2-p^2=\left(\frac{5}{7}-p\right)\left(\frac{5}{7}+p\right)$$

Ответ

$$x^2-y^2=(x-y)(x+y),\quad c^2-z^2=(c-z)(c+z),\quad a^2-25=(a-5)(a+5),$$

$$m^2-1=(m-1)(m+1),\quad 16-b^2=(4-b)(4+b),\quad 100-x^2=(10-x)(10+x),$$

$$p^2-400=(p-20)(p+20),\quad y^2-0{,}09=(y-0{,}3)(y+0{,}3),$$

$$1{,}44-a^2=(1{,}2-a)(1{,}2+a),\quad b^2-\frac{4}{9}=\left(b-\frac{2}{3}\right)\left(b+\frac{2}{3}\right),$$

$$\frac{9}{16}-n^2=\left(\frac{3}{4}-n\right)\left(\frac{3}{4}+n\right),\quad \frac{25}{49}-p^2=\left(\frac{5}{7}-p\right)\left(\frac{5}{7}+p\right).$$