Задачаа

[Гост]

20240420_155114.jpg (9.18 KiB) Прегледано 221 пъти

[ammornil]

[tex]\left(\frac{2x-3}{2} \right)^{2}-1\frac{2}{3}\left(\frac{x+1}{5}-\frac{3x+9}{10}\right)<(-x-1,5)^{2} \\ \phantom{q} \\ \frac{(2x-3)^{2}}{4}-\frac{5}{3}\left(\frac{2(x+1)-(3x+9)}{10}\right)<[-1(x+\frac{3}{2})]^{2} \\ \phantom{q} \\ \phantom{q} \\ \frac{(2x-3)^{2}}{4}-\frac{5}{3}\left(\frac{2x+2-3x-9}{10}\right)< \left(\frac{2x+3}{2}\right)^{2} \\ \phantom{q} \\ \frac{(2x-3)^{2}}{4}-\frac{5}{3}\left(-\frac{x+7}{10}\right)< \frac{(2x+3)^{2}}{4} \\ \phantom{q} \\ \underbrace{\frac{(2x-3)^{2}}{4}+\frac{5}{3}\cdot{}\frac{5(x+7)}{10}< \frac{(2x+3)^{2}}{4}}_{60} \\ \phantom{q} \\ 15(2x-3)^{2}-15(2x+3)^{2}+50(x+7)<0 \\ \phantom{q} \\ 15[(2x-3+2x+3)(2x-3-2x-3)]+50x+350<0 \\ \phantom{q} \\ 15\cdot{4x}\cdot{}(-6)+50x+350<0 \\ \phantom{q} \\ -360x+50x<-350 \\ \phantom{q} \\ x>\frac{350}{310} \\ \phantom{q} \\ x>1\frac{4}{31}[/tex]

[Евва]

Има техническа грешка в 5 ред .
Аз получих х>[tex]\frac{1}{5}[/tex]

[ammornil]

Благодаря, Евва. Ето коригираното решение.

[tex]\left(\frac{2x-3}{2} \right)^{2}-1\frac{2}{3}\left(\frac{x+1}{5}-\frac{3x+9}{10}\right)<(-x-1,5)^{2} \\ \phantom{q} \\ \frac{(2x-3)^{2}}{4}-\frac{5}{3}\left(\frac{2(x+1)-(3x+9)}{10}\right)<[-1(x+\frac{3}{2})]^{2} \\ \phantom{q} \\ \phantom{q} \\ \frac{(2x-3)^{2}}{4}-\frac{5}{3}\left(\frac{2x+2-3x-9}{10}\right)< \left(\frac{2x+3}{2}\right)^{2} \\ \phantom{q} \\ \frac{(2x-3)^{2}}{4}-\frac{5}{3}\left(-\frac{x+7}{10}\right)< \frac{(2x+3)^{2}}{4} \\ \phantom{q} \\ \underbrace{\frac{(2x-3)^{2}}{4}+\frac{5}{3}\cdot{}\frac{x+7}{10}< \frac{(2x+3)^{2}}{4}}_{60} \\ \phantom{q} \\ 15(2x-3)^{2}-15(2x+3)^{2}+10(x+7)<0 \\ \phantom{q} \\ 15[(2x-3+2x+3)(2x-3-2x-3)]+10x+70<0 \\ \phantom{q} \\ 15\cdot{4x}\cdot{}(-6)+10x+70<0 \\ \phantom{q} \\ -360x+10x<-70 \\ \phantom{q} \\ x>\frac{70}{350} \\ \phantom{q} \\ x>\frac{1}{5}[/tex]