Question

Реши уравнение: а)(5х-1)(5х+1)-(5х+2)^2=0 б)36b-121=0 хотя бы одно спасибо)

Answer 1

(5х-1)(5х+1)-(5х+2)²=0,
25х²-1-(25х²+20х+4)=0,
25х²-1-25х²-20х-4=0,
-5-20х=0,
-20х=5,
х=-0,25.
Ответ: -0,25.

36b-121=0,
36b=121,
b= $\frac{121}{36} =3 \frac{13}{36}$
Ответ: $3 \frac{13}{36}$

Answer 2

Solving the Equations

a) (5x-1)(5x+1)-(5x+2)^2=0

To solve this equation, we can start by expanding the terms and then simplifying the expression.

Expanding the terms: (5x-1)(5x+1) = 25x^2 - 1 (5x+2)^2 = 25x^2 + 20x + 4

Substitute the expanded terms back into the original equation: (25x^2 - 1) - (25x^2 + 20x + 4) = 0 25x^2 - 1 - 25x^2 - 20x - 4 = 0 -20x - 5 = 0 -20x = 5 x = -5/20 x = -1/4

So, the solution to the equation (5x-1)(5x+1)-(5x+2)^2=0 is x = -1/4.

b) 36b - 121 = 0

To solve this equation, we can isolate the variable b by adding 121 to both sides of the equation and then dividing by 36.

Adding 121 to both sides: 36b = 121

Dividing by 36: b = 121/36 b = 3.36

So, the solution to the equation 36b - 121 = 0 is b = 121/36 or 3.36.

I hope this helps! If you have any more questions or need further assistance, feel free to ask.