Question

Решите пожалуйста!! 1) (y+5)(y-1)=7 2) (3x-5)^+2x=5 3) t(t+2)-3(t-4)=5t+3

Answer 1

1
y²-y+5y-5-7=0
y²+4y-12=0
y1+y2=-4 U y1*y2=-12
y1=-6 U y2=2
2
9x²-30x+25+2x-5=0
9x²-28x+20=0
D=784-720=64
x1=(28-8)/18=10/9
x2=(28+8)/18=2
3
t²+2t-3t+12-5t-3=0
t²-6t+9=0
(t-3)²=0
t-3=0
t=3

Answer 2

1) (y+5)(y-1) = 7

To solve this equation, we can expand the left side of the equation and then set it equal to 7:

y^2 - y + 5y - 5 = 7

Simplifying the equation, we get:

y^2 + 4y - 12 = 0

Now, we can factor the quadratic equation:

(y + 6)(y - 2) = 0

Setting each factor equal to zero, we have:

y + 6 = 0 or y - 2 = 0

Solving for y, we get:

y = -6 or y = 2

Therefore, the solutions to the equation are y = -6 and y = 2.

2) (3x-5)^2x = 5

To solve this equation, we need to expand the left side of the equation and then solve for x:

(3x-5)^2 * x = 5

Expanding the equation, we get:

(9x^2 - 30x + 25) * x = 5

Multiplying through, we have:

9x^3 - 30x^2 + 25x = 5

Now, subtracting 5 from both sides of the equation, we get:

9x^3 - 30x^2 + 25x - 5 = 0

Unfortunately, this equation cannot be easily factored or solved algebraically. You may need to use numerical methods or a graphing calculator to find an approximate solution.

3) t(t+2) - 3(t-4) = 5t + 3

To solve this equation, we need to simplify and then solve for t:

t^2 + 2t - 3t + 12 = 5t + 3

Combining like terms, we get:

t^2 - 6t + 12 = 5t + 3

Subtracting 5t and 3 from both sides of the equation, we have:

t^2 - 11t + 9 = 0

Now, we can try to factor the quadratic equation:

(t - 9)(t - 1) = 0

Setting each factor equal to zero, we get:

t - 9 = 0 or t - 1 = 0

Solving for t, we have:

t = 9 or t = 1

Therefore, the solutions to the equation are t = 9 and t = 1.