Question

A)(x-6)^2-x^2=0 б)y(y-1)-(y-5)^2=2

Answer 1

Ответ:

$(x - 6) {}^{2} - x {}^{2} = 0\\ = x {}^{2} - 12x + 36 - x {}^{2} = 0 \\ - 12x + 36 = 0 \\ - 12x = - 36 \\ x = - 36 \div - 12 \\ x = 3$

Answer 2

Объяснение:

а) x²-12x+36-x²=0

-12x=-36

x=3

б)y²-y-y²+10y-25=2

9y=27

y=3

Answer 3

Let's solve these two equations separately:

A) Solve the equation $(x-6)^2 - x^2 = 0$:

Expand and simplify the equation:

$(x-6)^2 - x^2 = x^2 - 12x + 36 - x^2 = -12x + 36$

Now, set this expression equal to zero and solve for x:

$-12x + 36 = 0$

Subtract 36 from both sides:

$-12x = -36$

Divide both sides by -12:

$x = -36 / -12 = 3$

So, the solution to the equation $(x-6)^2 - x^2 = 0$ is $x = 3$.

B) Solve the equation $y(y-1) - (y-5)^2 = 2$:

Expand and simplify the equation:

$y(y-1) - (y-5)^2 = y^2 - y - (y^2 - 10y + 25) = y^2 - y - y^2 + 10y - 25 = -y + 10y - 25 = 9y - 25$

Now, set this expression equal to 2 and solve for y:

$9y - 25 = 2$

Add 25 to both sides:

$9y = 2 + 25 = 27$

Divide both sides by 9:

$y = 27 / 9 = 3$

So, the solution to the equation $y(y-1) - (y-5)^2 = 2$ is $y = 3$.