Question

дам 40 балов,срочно.Розкладіть на множники:
1) 16 — b°;
(5) 4х2 – 25;
2) с? — 49; 6) 81с2 – 64d”;
3) 0,04 – а”; 7) 0,09x? – 0,25y”;
8) a*b* – с®ds,
9) 4а?с? – 9x°у;
10) x24 - y22,
11) -1600 +a12;
12) a18 _ 49
64​

Answer 1

1. 16 — b²=(4-b)(4+b)

5. 4x²-25=(2x+5) (2x-5)

2. c²-49=(c-7)*(c+7)

3. 0.04-a²=(0.2-a)(0.2+a)

6. 81с² – 64d²=(9с-8d)(9с+8d)

7. 0,09x² – 0,25y²=(0.3x-0.5y)(0.3x+0.5y)

8. a²b²–с²d²=(ab–сd)(ab+сd)

9. 4а²с² – 9x²у²=(2ac-3xy)(2ac+3xy)

10. x²⁴- y²²=(x¹²-y¹¹)(x¹²+y¹¹)

11.  -1600 +a¹²=(a⁶-40)(a⁶+40)

12.  a¹⁸- 49=(a⁹-7)(a⁹+7)

Answer 2

$16-b^2=(4-b)(4+b)\\\\4x^2-25=(2x-5)(2x+5)\\\\c^2-49=(c-7)(c+7)\\\\81c^2-64d^2=(9c-8d)(9c+8d)\\\\0,04-a^2=(0,2-a)(0,2+a)\\\\0,09x^2-0,25y^2=(0,3x-0,5y)(0,3x+0,5y)\\\\a^2b^2-c^2d^2=(ab-cd)(ab+cd)\\\\4a^2c^2-9x^2y^2=(2ac-3xy)(2ac+3xy)\\\\x^{24}-y^{22}=(x^{12})^2-(y^{11})^2=(x^{12}-y^{11})(x^{12}+y^{11})\\\\-1600+a^{12}=(a^{6})^2-40^2=(a^6-40)(a^6+40)\\\\a^{18}-49=(a^9)^2-7^2=(a^9-7)(a^9+7)$