Задание 7 Параграф 19 ГДЗ Рабочая тетрадь 1 Мерзляк Полонский 7 класс (Алгебра)

1) x^2+5x-6;
2) a^2-10a+16;
3) y^2+y-2;
4) 4b^2-11b-3.

1. $$x^2+5x-6=x^2+2\cdot x\cdot 2{,}5+2{,}5^2-2{,}5^2-6$$
   $$=(x+2{,}5)^2-12{,}25=(x+2{,}5)^2-3{,}5^2$$
   $$=(x+2{,}5-3{,}5)(x+2{,}5+3{,}5)=(x-1)(x+6).$$
2. $$a^2-10a+16=a^2-2\cdot a\cdot 5+5^2-5^2+16$$
   $$=(a-5)^2-9=(a-5)^2-3^2$$
   $$=(a-5-3)(a-5+3)=(a-8)(a-2).$$
3. $$y^2+y-2=y^2+2\cdot y\cdot \frac12+\left(\frac12\right)^2-\left(\frac12\right)^2-2$$
   $$=\left(y+\frac12\right)^2-\frac14-2=\left(y+\frac12\right)^2-\frac94$$
   $$=\left(y+\frac12-\frac32\right)\left(y+\frac12+\frac32\right)=(y-1)(y+2).$$
4. $$4b^2-11b-3=4b^2-2\cdot 2b\cdot \frac{11}{4}+\left(\frac{11}{4}\right)^2-\left(\frac{11}{4}\right)^2-3$$
   $$=\left(2b-\frac{11}{4}\right)^2-\frac{121}{16}-3=\left(2b-\frac{11}{4}\right)^2-\frac{169}{16}$$
   $$=\left(2b-\frac{11}{4}-\frac{13}{4}\right)\left(2b-\frac{11}{4}+\frac{13}{4}\right)=(2b-6)(2b+0{,}5)$$
   $$=2\cdot 0{,}5\,(b-3)(4b+1)=(b-3)(4b+1).$$

Ответ

1) $$x^2+5x-6=(x-1)(x+6)$$
2) $$a^2-10a+16=(a-8)(a-2)$$
3) $$y^2+y-2=(y-1)(y+2)$$
4) $$4b^2-11b-3=(b-3)(4b+1)$$