Задача

[Гост]

IMG-c01d4524cc77302c68ecc4d199cf8bc4-V.jpg (29.16 KiB) Прегледано 426 пъти

[ammornil]

Нещо такова може би...
[tex]\left( \underbrace{\overset{2}{1}-\frac{x}{2}}_{2} \right)^{2}-\frac{3}{2}\cdot \left( 4+\frac{3x-4}{3} \right)=\frac{x^{2}+1}{4}-\frac{1}{3} \Leftrightarrow \left(\frac{2+x}{2}\right)^{2}-\frac{3}{\cancel{2}}\cdot{\overset{\normalsize{2}\hspace{1em}}{\cancel{4}}}-\frac{\cancel{3}}{2}\cdot{\frac{3x-4}{\cancel{3}}}=\frac{x^{2}+1}{4}-\frac{1}{3} \Leftrightarrow \\ \phantom{q} \\ \Leftrightarrow \underbrace{\frac{\overset{3}{(2+x)^{2}}}{4}-\overset{12}{6}-\frac{\overset{6}{3x-4}}{2}=\frac{\overset{3}{x^{2}+1}}{4}-\frac{\overset{4}{1}}{3}}_{12} \Leftrightarrow 3\cdot (4+4x+x^{2})-72-6\cdot (3x-4)=3\cdot (x^{2}+1)-4 \Leftrightarrow \\ \phantom{q} \\ \Leftrightarrow \blue{12}\red{+12x}+\cancel{3x^{2}}\blue{-72}\red{-18x}\blue{+24}=\cancel{3x^{2}}\purple{+3-4 }\Leftrightarrow -6x=-1+36[/tex]$$ x=-\frac{35}{6}=-5\frac{5}{6} $$