Упр.24.27 ГДЗ Мерзляк Поляков 8 класс (Алгебра)

1) x^2+(x/(x-1))^2=8; 3) (x/(x-1))^2+(x/(x+1))^2=90;
2) x^2+(x/(2x-1))^2=2; 4) x^2+(25x^2)/(5+2x)^2 =104.

1. $$x^2+\left(\frac{x}{x-1}\right)^2=8, \qquad x\ne 1$$

   Преобразуем:

   $$ x^2+2x\cdot \frac{x}{x-1}+\left(\frac{x}{x-1}\right)^2-2x\cdot \frac{x}{x-1}=8 $$
   $$ \left(x+\frac{x}{x-1}\right)^2-\frac{2x^2}{x-1}-8=0 $$
   $$ \left(\frac{x(x-1)+x}{x-1}\right)^2-\frac{2x^2}{x-1}-8=0 $$
   $$ \left(\frac{x^2}{x-1}\right)^2-\frac{2x^2}{x-1}-8=0 $$

   Пусть $$\frac{x^2}{x-1}=t$$, тогда получаем:

   $$t^2-2t-8=0$$

   $$t_1=4,\quad t_2=-2$$

   Решаем два уравнения:

   $$ \frac{x^2}{x-1}=4 \Rightarrow x^2=4x-4 \Rightarrow x^2-4x+4=0 \Rightarrow (x-2)^2=0 \Rightarrow x=2 $$
   $$ \frac{x^2}{x-1}=-2 \Rightarrow x^2=-2x+2 \Rightarrow x^2+2x-2=0 $$
   $$ D=4+8=12,\quad x=\frac{-2\pm \sqrt{12}}{2}=-1\pm \sqrt{3} $$

2. $$x^2+\left(\frac{x}{2x-1}\right)^2=2, \qquad x\ne \frac12$$

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

   Пусть $$\frac{2x^2}{2x-1}=t$$. Тогда:

   $$t^2-t-2=0$$

   $$t_1=2,\quad t_2=-1$$

   $$ \frac{2x^2}{2x-1}=2 \Rightarrow 2x^2=4x-2 \Rightarrow x^2-2x+1=0 \Rightarrow x=1 $$
   $$ \frac{2x^2}{2x-1}=-1 \Rightarrow 2x^2=-2x+1 \Rightarrow 2x^2+2x-1=0 $$
   $$ D=4+8=12,\quad x=\frac{-2\pm \sqrt{12}}{4}=\frac{-1\pm \sqrt{3}}{2} $$

3. $$\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=90, \qquad x\ne \pm 1$$

   $$ \left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2-\frac{2x^2}{x^2-1}-90=0 $$
   $$ \left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-90=0 $$

   Пусть $$\frac{2x^2}{x^2-1}=t$$. Тогда:

   $$t^2-t-90=0$$

   $$t_1=10,\quad t_2=-9$$

   $$ \frac{2x^2}{x^2-1}=10 \Rightarrow 2x^2=10x^2-10 \Rightarrow 8x^2=10 \Rightarrow x^2=\frac54 \Rightarrow x=\pm \frac{\sqrt5}{2} $$
   $$ \frac{2x^2}{x^2-1}=-9 \Rightarrow 2x^2=-9x^2+9 \Rightarrow 11x^2=9 \Rightarrow x^2=\frac{9}{11} \Rightarrow x=\pm \frac{3\sqrt{11}}{11} $$

4. $$x^2+\frac{25x^2}{(5+2x)^2}=104, \qquad x\ne -\frac52$$

   $$ \left(x-\frac{5x}{5+2x}\right)^2+\frac{10x^2}{5+2x}-104=0 $$
   $$ \left(\frac{2x^2}{5+2x}\right)^2+\frac{10x^2}{5+2x}-104=0 $$

   Пусть $$\frac{2x^2}{5+2x}=t$$. Тогда:

   $$t^2+5t-104=0$$

   $$t_1=8,\quad t_2=-13$$

   $$ \frac{2x^2}{5+2x}=8 \Rightarrow 2x^2=40+16x \Rightarrow x^2-8x-20=0 $$
   $$ D=64+80=144,\quad x=\frac{8\pm 12}{2} $$
   $$ x_1=10,\quad x_2=-2 $$
   $$ \frac{2x^2}{5+2x}=-13 \Rightarrow 2x^2=-65-26x \Rightarrow 2x^2+26x+65=0 $$
   $$ D=26^2-4\cdot 2\cdot 65=156,\quad x=\frac{-26\pm \sqrt{156}}{4}=\frac{-13\pm \sqrt{39}}{2} $$

Ответ

1) $$x=-1\pm \sqrt3,\; 2$$

2) $$x=1,\; \frac{-1\pm \sqrt3}{2}$$

3) $$x=\pm \frac{3\sqrt{11}}{11},\; \pm \frac{\sqrt5}{2}$$

4) $$x=-2,\; 10,\; \frac{-13\pm \sqrt{39}}{2}$$