Как мога да подходя тук?

[Wiktor]

[tex]\frac{x+1}{x-1}+ \frac{x-2}{x+2}+ \frac{x-3}{x+3}+ \frac{x+4}{x-4} = 4[/tex]

[ammornil]

[tex]\frac{x+1}{x-1}+ \frac{x-2}{x+2}+ \frac{x-3}{x+3}+ \frac{x+4}{x-4} = 4[/tex]

ДМ [tex]\begin{array}{|l} x-1 \ne 0 \\ x+2 \ne 0 \\ x+3 \ne 0 \\ x-4 \ne 0 \end{array} \Leftrightarrow \begin{array}{|l} x \ne 1 \\ x \ne -2 \\ x \ne -3 \\ x \ne 4 \end{array} \Rightarrow x \in (- \infty;-3) \cup (-3;-2) \cup (-2;1) \cup (1;4) \cup (4;+ \infty )[/tex]

Можем да си спестим четвъртите степени, но това е всичко което аз мога да се сетя. Прегледай сметките и най-вече знаците на преобразуванията.

[tex]\frac{x-1+2}{x-1}+ \frac{x+2-4}{x+2}+ \frac{x+3-6}{x+3}+ \frac{x-4+8}{x-4} = 4 \Leftrightarrow \frac{x-1}{x-1}+\frac{2}{x-1}+ \frac{x+2}{x+2}- \frac{4}{x+2}+ \frac{x+3}{x+3}- \frac{6}{x+3}+\frac{x-4}{x-4}+\frac{8}{x-4}=4 \Leftrightarrow[/tex]

[tex]\Leftrightarrow 1+\frac{2}{x-1}+ 1- \frac{4}{x+2}+1- \frac{6}{x+3}+1+\frac{8}{x-4}=4 \Leftrightarrow \underbrace{\frac{2}{x-1}- \frac{4}{x+2}- \frac{6}{x+3}+\frac{8}{x-4}=0}_{(x-1)(x+2)(x+3)(x-4)} \Leftrightarrow[/tex]

[tex]\Leftrightarrow \underline{2(x+2)(x+3)(x-4)-4(x-1)(x+3)(x-4)}-6(x-1)(x+2)(x-4)+8(x-1)(x+2)(x+3)=0 \Leftrightarrow[/tex]

[tex]\Leftrightarrow (x+3)(x-4)[2(x+2)-4(x-1)]-(x-1)(x+2)[6(x-4)-8(x+3)]=0 \Leftrightarrow[/tex]

[tex]\Leftrightarrow (x+3)(x-4)(2x+4-4x+4)-(x-1)(x+2)(6x-24-8x-24)=0 \Leftrightarrow -2(x-4)(x+3)(x-4)+2(x-1)(x+2)(x+24)=0 \Leftrightarrow[/tex]

[tex](-2x-6)(x^{2}-8x+16)+(2x+48)(x^{2}+x-2)=0 \Leftrightarrow -2x^{3}+16x^{2}-32x-6x^{2}+48x-96+2x^{3}+2x^{2}-4x+48x^{2}+48x-96=0 \Leftrightarrow[/tex]

[tex]\Leftrightarrow 60x^{2}+60x-192=0 |:12 \Leftrightarrow 5x^{2}+5x-16=0 \rightarrow D=5^{2}-4.5.(-16)=345 \rightarrow x_{_{1,2}}=\frac{-5\pm\sqrt{345}}{10}[/tex]