Question

Вычислите:
a) 97^2-94*97+47^2
b)53^2-27^2/79^2-51^2
2.Решите уравнение:
(4x+3)^2-(4x-5)(4x+5)=82

Answer 1

$a) \ 97^2-94\cdot 97+47^2 = 97^2-2\cdot 47\cdot 97+47^2 = (97-47)^2=50^2=2500; \\ \\ b) \ \frac{53^2-27^2}{79^2-51^2}=\frac{(53-27)\cdpt(53+27)}{(79-51)\cdot(79+51)}=\frac{26\cdot 80}{28\cdot 130}=\frac{26\cdot 8}{28\cdot 13}=\frac{2\cdot8}{28\cdot1}=\frac{8}{14}=\frac{4}{7}$

$2. \ (4x+3)^2-(4x-5)\cdot(4x+5)=82; \\ (4x)^2+2\cdot4x\cdot 3+3^2 -((4x)^2-5^2)=82; \\ 16x^2+24x+9-(16x^2-25)=82; \\ 16x^2+24x+9-16x^2+25=82; \\ 24x+9+25=82; \\ 24x=82-9-25; \\ 24x=48; \\ \\ x=\frac{48}{24}=2$