Question

Решите дробно рациональные уравнения х^2/2x+3 - x/2 = 0 и 2x-5/x+5 + 3x+4/x+2 = 1

Answer 1

$1)\frac{x^{2} }{2x+3} -\frac{x}{2}=0\\\\\frac{2x^{2}-2x^{2}-3x}{2(2x-3)} =0\\\\\frac{-3x}{2x-3}=0\\\\\left \{ {{-3x=0} \atop {2x-3\neq 0}} \right.\\\\\left \{ {{x=0} \atop {x\neq 1,5}} \right.$

ответ : 0

$2)\frac{2x-5}{x+5}+\frac{3x+4}{x+2}=1\\\\\frac{2x^{2}+4x-5x-10+3x^{2}+4x+15x+20}{(x+5)(x+2)} =1\\\\\frac{5x^{2}+18x+10 }{(x+5)(x+2)}=1 \\\\\left \{ {{5x^{2}+18x+10-x^{2}-2x-5x-10=0} \atop {x+5\neq0;x+2\neq0 }} \right.\\\\\left \{ {{4x^{2}+11x=0 } \atop {x\neq-5;x\neq -2 }} \right. \\\\4x^{2}+11x=0\\\\x(4x+11)=0\\\\x_{1}=0\\\\4x+11=0\\\\x_{2} =-2,75$

Answer 2

1) $\(\frac{x^2}{2x+3}-\frac{x}{2}=0\)$

ОДЗ: x ≠ -1,5

$\(\frac{x^2}{2x+3}-\frac{x}{2}=0|\cdot2(2x+3)\)$

$\(2x^2-x(2x+3)=0\)$

2x² - 2x² - 3x = 0;

-3x = 0;

x = 0.

ответ: 0.

2) $\(\frac{2x-5}{x+5}+\frac{3x+4}{x+2}=1\)$

ОДЗ: x ≠ -5; x ≠ -2.

$\(\frac{2x-5}{x+5}+\frac{3x+4}{x+2}=1|\cdot(x+5)(x+2)\)$

(2x - 5)(x + 2) + (3x + 4)(x + 5) = (x + 5)(x + 2).

2x² - x - 10 + 3x² + 19x + 20 = x² + 7x + 10;

4x² + 11x  = 0;

x² + 2,75x = 0.

x₁ = 0; x₂ = -2,75

ответ: 0; -2,75.