Question

1.408,1.410,1.411 под буквами "а"

Answer 1

Решение во вложенииииииииииииииииииииии
11 добавлю
добавила

Answer 2

1.408a
$f(x)= \frac{1}{x+1} + \frac{2}{x-1} \\\ f(x)=f(a), \ a=3 \\\ \frac{1}{x+1} + \frac{2}{x-1} =\frac{1}{3+1} + \frac{2}{3-1} \\\ \frac{1}{x+1} + \frac{2}{x-1} =\frac{1}{4} + \frac{2}{2} \\\ \frac{1}{x+1} + \frac{2}{x-1} =\frac{5}{4} \ (x \neq \pm1) \\\ 4(x-1)+8(x+1)=5(x^2-1) \\\ 4x-4+8x+8=5x^2-5 \\\ 5x^2-12x-9=0 \\\ D_1=(-6)^2-5\cdot(-9)=36+45=81 \\\ x_1= \frac{6+9 }{5} =3 \\\ x_2= \frac{6-9 }{5} =-0.6$
Ответ: 3 и -0,6

1.410а
$\phi(x)=x-3+ \frac{1-4x}{x} =x-3+ \frac{1}{x} -4=x+ \frac{1}{x} -7, \ x \neq 0 \\\ \phi(2.5)=2.5+ \frac{1}{2.5} -7=-4.5+ \frac{1}{5/2}=-4.5+ \frac{2}{5}=-4.5+0.4=-4.1 \\\ \phi(0.4)=0.4+ \frac{1}{0.4} -7=-6.6+ \frac{1}{2/5}=-6.6+ \frac{5}{2}=-6.6+2.5=-4.1 \\\ \Rightarrow \phi(2.5)=\phi(0.4)$

1.411a
$y_1=x^2+ \frac{1}{x-4} , \ x \neq -4 \\\ y_2= \frac{4x-15}{x-4} = \frac{4x-16+1}{x-4} =\frac{4(x-4)+1}{x-4} =4+\frac{1}{x-4} , \ x \neq -4 \\\ y_1=y_2 \\\ x^2+ \frac{1}{x-4} =4+ \frac{1}{x-4} \\\ x^2=4 \\\ x=\pm 2$