Здравейте, имам задачи за домашно...

[Гост]

Моля помогнете ми за утре е !

[mail_dinko]

[tex]\frac {(x-1)^2}{4}-2x=(\frac{x-3}{2})^2-2|.4[/tex]
[tex](x-1)^2-8x-(x-3)^2+8=0[/tex]
[tex]\cancel {x^2}-2x\cancel{+1}-8x\cancel {-x^2}+6x\cancel{-9+8}=0[/tex]
[tex]-4x=0[/tex]
[tex]x=0[/tex]

[tex]\frac{2x^2-x-3}{6x^2+3x-18}=\frac{2x^2-x-3}{3(2x^2+x-6)}[/tex]
[tex]2x^2-x-3=0[/tex]
[tex]D=1+24=25=5^2[/tex]
[tex]x_{1,2}=\frac {1 \pm 5}{2.2}[/tex]
[tex]x_1 = \frac {3}{2}[/tex]
[tex]x_2 = -1[/tex]
[tex]2x^2-x-3=2(x+1)(x- \frac32)=[/tex]
[tex]2x^2+x-6=0[/tex]
[tex]D=1+48=49=7^2[/tex]
[tex]x_{3,4}= \frac {-1 \pm 7}{2.2}[/tex]
[tex]x_3 = -2[/tex]
[tex]x_4 = \frac {3}{2}[/tex]
[tex]2x^2+x-6=2(x+2)(x- \frac {3}{2})[/tex]
[tex]\frac{2x^2-x-3}{3(2x^2+x-6)} = \frac {\cancel {2}(x+1)\cancel {(x- \frac32)}}{\cancel {2}(x+2)\cancel {(x- \frac {3}{2})}}= \frac {x+1}{x+2}[/tex]