Задачи

[Гост]

Моля за помощ на тези задачи:
Зад1)

20250403_144013.jpg (601.35 KiB) Прегледано 170 пъти

Зад2)

20250403_144108.jpg (426.91 KiB) Прегледано 170 пъти

[ammornil]

$\boxed{1}\\[6pt] \dfrac{x}{3-x}-\dfrac{x+3}{1-3x}= \dfrac{x^{2}-4x-7}{3x^{2} -10x +3} \\[6pt] \quad 3x^{2} -10x +3 \rightarrow x_{1,2}=\dfrac{-(-10)\pm\sqrt{(-10)^{2} -4\cdot{}3\cdot{}3}}{2\cdot{}3}= \dfrac{10\pm{8}}{6} \\[6pt] \quad \Rightarrow x_{1}=\dfrac{1}{3}, \hspace{0.4em} x_{2}=3 \Rightarrow \\[6pt] \quad 3x^{2} -10x +3 =3(x-3)\left(x-\dfrac{1}{3}\right)=(x-3)(3x-1) \\[6pt] \dfrac{x}{-(x-3)}-\dfrac{x+3}{-(3x-1)}= \dfrac{x^{2}-4x-7}{(x-3)(3x-1)} \\[6pt] \text{ДМ: } \quad \begin{cases} x-3\ne{0} \\ 3x-1\ne{0} \end{cases} \Rightarrow x\ne{3} \cup x\ne{\dfrac{1}{3}} \\[6pt] \underbrace{\dfrac{\overset{3x-1}{-x}}{x-3} +\dfrac{\overset{x-3}{x+3}}{3x-1} =\dfrac{\overset{1}{x^{2} -4x -7}}{(x-3)(3x-1)}}_{(x-3)(3x-1)} \\[6pt] -x(3x-1) +(x+3)(x-3)=x^{2}-4x-7 \\[6pt] -3x^{2} +x +x^{2} -9 -x^{2} +4x +7 =0 \\[6pt] -3x^{2} +5x -2=0 \quad |\cdot{}(-1) \\[6pt] 3x^{2} -5x +2=0 \\[6pt] \quad x_{1,2}=\dfrac{-(-5)\pm\sqrt{(-5)^{2} -4\cdot{}3\cdot{}2}}{2\cdot{}3} =\dfrac{5\pm{1}}{6} \\[6pt] x_{1}=-\dfrac{2}{3}\in\text{ДМ} \quad \cup \quad x_{2}=1\in\text{ДМ} $

[ammornil]

$\dfrac{3x-1}{2x-3}-\dfrac{x-2}{2+x}= \dfrac{2x+3}{2x^{2}+x-6} \\[6pt] \quad 2x^{2}+x-6 \rightarrow x_{1,2}=\dfrac{-1\pm\sqrt{1^{2}-4\cdot{}2\cdot{}(-6)}}{2\cdot{}2} \Rightarrow x_{1,2}=\dfrac{-1\pm{}7}{4} \\[6pt] \quad \Rightarrow x_{1}=-2, \quad x_{2}=\dfrac{3}{2} \\[6pt] \quad 2x^{2}+x-6= 2(x+2)\left(x-\dfrac{3}{2}\right)= (x+2)(2x-3) \\[6pt] \dfrac{3x-1}{2x-3}-\dfrac{x-2}{2+x}= \dfrac{\quad2\left(x+\dfrac{3}{2}\right)\quad}{(x +2)(2x -3)} \\[6pt] \quad \text{ДМ: } \begin{cases} 2x-3\ne{}0 \\ x+2\ne{}0 \\ (x+2)\left(2x-3\right) \ne{} 0 \end{cases} \Rightarrow x\ne{\dfrac{3}{2}} \cup x\ne{-2} \\[12pt] \underbrace{ \dfrac{\overset{x+2}{3x-1}}{2x-3}-\dfrac{\overset{2x-3}{x-2}}{2+x}= \dfrac{\overset{1}{2x+3}}{2x^{2}+x-6}}_{(x+2)(2x -3)} \\[6pt] (x+2)(3x-1)-(2x-3)(x-2)=2x+3 \\[6pt] 3x^{2} -x +6x -2 -(2x^{2} -4x -3x +6) -2x -3=0 \\[6pt] 3x^{2} -x +6x -2 -2x^{2} +7x -6 -2x -3=0 \\[6pt] x^{2} +10x -11=0 \\[6pt] \quad x_{1,2}= \dfrac{-10\pm\sqrt{10^{2}-4\cdot{}1\cdot{}(-11)}}{2\cdot{}1}= \dfrac{-10\pm12}{2} \\[6pt] x_{1}=-11 \cup x_{2}=1$