Question

(х-5)в квадрате=5х в квадрате-(2х-1)(2х+1)

Answer 1

$(x-5)^{2}=5x^{2}-(2x-1)(2x+1)$

 

$x^{2}-10x+25=5x^{2}-(4x^{2}-1)$

 

$x^{2}-10x+25=5x^{2}-4x^{2}+1$

 

$x^{2}-10x+25=x^{2}+1$

 

$x^{2}+1-x^{2}+10x-25=0$

 

$(x^{2}-x^{2})+10x+(1-25)=0$

 

$10x-24=0$

 

$10x=24$

 

$x=24:10$

 

$x=2,4$

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$(2,4-5)^{2}=5\cdot(2,4)^{2}-(2\cdot2,4-1)\cdot(2\cdot2,4+1)$ (это проверка)

 

$(-2,6)^{2}=5\cdot5,76-(4,8-1)\cdot(4,8+1)$

 

$6,76=28,8-3,8\cdot5,8$

 

$6,76=28,8-22,04$

 

$6,76=6,76$

Answer 2

x^2 - 10x +25= 5x^2 - 4x^2 - 2x +2x +1

x^2 - 10x +25= x^2 +1

-10x+25=1

-10x=-24

x=24/10

x=12/5

x=2,4