Решите плз (2x^2+3x)/(3-x)=(x-x^2)/(x-3) (5x-7)/(x-3)=(4x-3)/x

Question

Решите плз
(2x^2+3x)/(3-x)=(x-x^2)/(x-3)
(5x-7)/(x-3)=(4x-3)/x

Answer 1

(2x²+3x)/(3-x)=(x-x²)/(x-3), x≠3
(2x²+3x)/(3-x)-(x-x²)/(x-3)=0
(2x²+3x)/(3-x)+(x-x²)/(3-x)=0 |*(3-x)
2x²+3x+x-x²=0
x²+4x=0
x(x+4)=0
x=0
x=-4

Ответ: 0; -4

(5x-7)/(x-3)=(4x-3)/x |*x(x-3)
x(5x-7)-(x-3)(4x-3)=0
5x²-7x-(4x²-3x-12x+9)=0
5x²-7x-4x²+3x+12x-9=0
x²+8x-9=0
a=1, b=8, c=-9
D/4 = (b/2)²-ac
D/4 = 16+9 = 25, √D/4 = 5
x= (-b/2+-√D/4)/a
x₁ = -4+5 = 1
x₂ = -4-5 = -9

Ответ: 1;-9

Comments

Второй решишь?

---

сейчас, я про него забыл

---

ок

---

все, обнови страницу

---

Пасиб

Answer 2

1
(2x²+3x)/(3-x)=(x²-x)/(3-x)
x≠3
2x²+3x=x²-x
x²+4x=0
x(x+4)=0
x1=0
x2=-4
2
(5x-7)/(x-3)-(4x-3)/x=0
x≠0 x≠3
(5x-7)x-(4x-3)(x-3)=0
5x²-7x-4x²+12x+3x-9=0
x²+8x-9=0
x1+x2=-8 U x1*x2=-9
x1=-9
x2=1