Question

3х-х (в кводрате)+2х(в кводрате)=х                         2х(в кводрате)+3х=х-х(в кводрате)
---------------------           -----------------                              ----------------           ---------------
             2                          2                                            3-х                     3-х

№3
5х-7=4х-3                                                       №4         2х+3=3х+2
------  ------                                                                       -------  -------
х-3      х                                                                          х+2    х

№5
4   - 5 = 1
----  -----  -----                                        №6        5y+13  -  4-6y  =  3
x+3  3-x x-3                                                     ----------      ------   
                                                                        5y+4        3y-1   
 


 
                                               

Answer 1

1) $frac{3x-x^2}{2}+ frac{2x^2}{2}=x$
$frac{3x-x^2+2x^2-2x}{2}=0$
$x^2+x=0$
$x*(x+1)=0$
$x_1=0$
$x+1=0$
$x_2=-1$

2) $frac{2x^2+3x}{3-x} = frac{x-x^2}{3-x}$
$frac{2x^2+3x}{3-x} - frac{x-x^2}{3-x}=0$
$2x^2+3x-x+x^2=0$
$3x^2+2x=0$
$x*(3x+2)$
$x_1=0$
$3x+2=0$
$3x=-2$
$x_2= -frac{2}{3}$

3) $frac{5x-7}{x-3} = frac{4x-3}{x}$
$frac{5x-7}{x-3} - frac{4x-3}{x} =0$
$5x^2-7x-4x^2+3x+12x-9=0$
$x^2+8x-9=0$
по теореме Виета $x_1=-9$
                            $x_2 = 1$

4) $frac{2x+3}{x+2}= frac{3x+2}{x}$
$frac{2x+3}{x+2}-frac{3x+2}{x}=0$
$2x^2+3x-3x^2-2x-6x-4=0$
$-x^2-5x-4=0$
$x^2+5x+4=0$
По теореме Виета $x_1 = -4$
                            $x_2=-1$

5) $frac{4}{x+3}- frac{5}{3-x} = frac{1}{x-3}$
$frac{4}{x+3}- frac{5}{3-x} - frac{1}{x-3}=0$
$4x-12+5x+15-x-3=0$
$8x=0$
$x=0$

6) $frac{5y+13}{5y+4}- frac{4-6y}{3y-1}=3$
$frac{5y+13}{5y+4}- frac{4-6y}{3y-1}-3 =0$
$15y^2+39y-5y-13-20y+30y^2-16+24y-45y^2-36y+15y+12=0$
$17y=-17$
$y=-1$