Question

Решите уравнения.
(Ответы -2 и -3,75 не подходят)

Answer 1

$\displaystyle 1.\\\\\dfrac x{x+3}+\dfrac{4x+6}{x^2+4x+3}=0\\\\\\\dfrac x{x+3}^{(x+1}+\dfrac{4x+6}{(x+3)(x+1)}=0\\\\\\\dfrac{x^2+x+4x+6}{(x+3)(x+1)}=0\ \ \ \ \ \Leftrightarrow\ \ \ \ \dfrac{x^2+5x+6}{(x+3)(x+1)}=0\\\\\\\dfrac{(x+3)(x+2)}{(x+3)(x+1)}=0\ \ \ \ \Leftrightarrow\ \ \ \left \{ {{(x+3)(x+2)=0} \atop {(x+3)(x+1)\neq0}} \right. \\\\\\\left \{ {{\left [ {{x=-3} \atop {x=-2}} \right. } \atop {x\neq-3;\ x\neq-1}} \right.\ \ \ \Rightarrow \ \ \ \boxed{\boldsymbol{x=-2}}$

Проверка :

$\dfrac{-2}{-2+3}+\dfrac{4\cdot(-2)+6}{(-2)^2+4\cdot(-2)+3}=\\\\=\dfrac{-2}1+\dfrac{-8+6}{4-8+3}=-2+\dfrac{-2}{-1}=-2+2\boldsymbol{=0}$

Ответ : -2  -  проверка подтвердила

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$2.\\\\\dfrac{x^2}{x-3}-\dfrac{45}{x^2-x-6}+\dfrac x{x+2}=x\\\\\\\dfrac{x^2}{x-3}^{(x+2}-\dfrac{45}{(x-3)(x+2)}+\dfrac x{x+2}^{(x-3}-\Big x^{(x^2-x-6}=0\\\\\\\dfrac{x^2(x+2)-45+x(x-3)-x(x^2-x-6)}{(x-3)(x+2)}=0\\\\\\\dfrac{\underline{x^3}+\underline{\underline{2x^2}}-45+\underline{\underline{x^2}}-\underline{\underline{\underline{3x}}}-\underline{x^3}+\underline{\underline{x^2}}+\underline{\underline{\underline{6x}}}}{(x-3)(x+2)}=0$

$\displaystyle\dfrac{4x^2+3x-45}{(x-3)(x+2)}=0\ \ \ \Leftrightarrow\ \ \ \left \{ {{4x^2+3x-45=0} \atop {(x-3)(x+2)\neq0}} \right.$

$4x^2+3x-45=0\\D=3^2-4\cdot 4\cdot(-45)=729=27^2\\\\x_1=\dfrac{-3+27}{2\cdot 4}=\dfrac{24}8=3\\\\x_2=\dfrac{-3-27}{2\cdot 4}=\dfrac{-30}{8}=-3\dfrac34=-3,75$

$\displaystyle \left \{ {{\left [ {{x_1=3~~~~~} \atop {x_2=-3,75}} \right. } \atop {x\neq3;x\neq-2}} \right. \ \ \ \ \ \Rightarrow\ \ \ \ \boxed{\boldsymbol{x=-3\dfrac 34=-3,75}}$

Проверка :

$\dfrac{\big(-3\frac 34\big)^2}{-3\frac 34-3}-\dfrac{45}{\big(-3\frac 34\big)^2-\big(-3\frac 34\big)-6}+\dfrac {-3\frac 34}{-3\frac34+2}=\\\\\\=\dfrac{\big(-\frac {15}4\big)^2}{-6\frac 34}-\dfrac{45}{\big(-\frac {15}4\big)^2+3\frac 34-6}+\dfrac {-3\frac 34}{-1\frac34}=\\\\\\=\dfrac {225}{16}:\bigg(-\dfrac {27}4\bigg)-\dfrac{45}{\frac {225}{16}+3\frac 34-6}+\bigg(-\dfrac {15}4\bigg):\bigg(-\dfrac74}\bigg)=$

$=-\dfrac {225}{16}\cdot \dfrac 4{27}-\dfrac{45}{14\frac 1{16}+3\frac {12}{16}-6}+\dfrac {15}4\cdot\dfrac47=\\\\\\=-\dfrac {9\cdot 25\cdot 4}{4\cdot 4\cdot 9\cdot 3}-\dfrac{45}{11\frac {13}{16}}+\dfrac {15}7=-\dfrac{25}{12}-45:\dfrac{189}{16}+2\dfrac 17=\\\\\\=-2\dfrac1{12}-45\cdot \dfrac{16}{189}+2\dfrac 17=-2\dfrac1{12}- \dfrac{9\cdot 5\cdot 16}{9\cdot 21}+2\dfrac 17=\\\\\\=-2\dfrac1{12}- \dfrac{80}{21}+2\dfrac 17^{(3}=-2\dfrac1{12}- 3\dfrac{17}{21}+2\dfrac 3{21}=$

$=-2\dfrac1{12}-1\dfrac{14}{21}=-2\dfrac1{12}-1\dfrac23^{(4}=\\\\\\=-2\dfrac1{12}-1\dfrac8{12}=-3\dfrac 9{12}\boldsymbol{=-3\dfrac 34=-3,75}$

Ответ :  -3,75  -  проверка подтвердила