Решить неравенство -3х квадрат +х -3 меньше или равно нулю
-3x²+x-3≤0 3x²-x+3≥0 D=1-36= -35 D<0 следовательно решений нет.<br>x∈R (х принадлежит всей числовой прямой)
![3*(x^2-\frac{5}{3}x)+3 \geq 0\\\\ 3*(x^2-2*x*\frac{5}{3*2})+3 \geq 0\\\\ 3*[x^2-2*x*\frac{5}{6}+(\frac{5}{6})^2-(\frac{5}{6})^2]+3 \geq 0\\\\ 3*[(x-\frac{5}{6})-(\frac{5}{6})^2]+3 \geq 0\\\\ 3*(x-\frac{5}{6})^2-3*(\frac{5}{6})^2+3 \geq 0\\\\ 3*(x-\frac{5}{6})^2-3*\frac{25}{36}+3 \geq 0\\\\ 3*(x-\frac{5}{6})^2-\frac{25}{12}+\frac{36}{12} \geq 0\\\\ 3*(x-\frac{5}{6})^2+\frac{11}{12} \geq 0\\\\ x\in(-\infty;\ +\infty)](https://tex.z-dn.net/?f=3%2A%28x%5E2-%5Cfrac%7B5%7D%7B3%7Dx%29%2B3+%5Cgeq+0%5C%5C%5C%5C+3%2A%28x%5E2-2%2Ax%2A%5Cfrac%7B5%7D%7B3%2A2%7D%29%2B3+%5Cgeq+0%5C%5C%5C%5C+3%2A%5Bx%5E2-2%2Ax%2A%5Cfrac%7B5%7D%7B6%7D%2B%28%5Cfrac%7B5%7D%7B6%7D%29%5E2-%28%5Cfrac%7B5%7D%7B6%7D%29%5E2%5D%2B3+%5Cgeq+0%5C%5C%5C%5C+3%2A%5B%28x-%5Cfrac%7B5%7D%7B6%7D%29-%28%5Cfrac%7B5%7D%7B6%7D%29%5E2%5D%2B3+%5Cgeq+0%5C%5C%5C%5C+3%2A%28x-%5Cfrac%7B5%7D%7B6%7D%29%5E2-3%2A%28%5Cfrac%7B5%7D%7B6%7D%29%5E2%2B3+%5Cgeq+0%5C%5C%5C%5C+3%2A%28x-%5Cfrac%7B5%7D%7B6%7D%29%5E2-3%2A%5Cfrac%7B25%7D%7B36%7D%2B3+%5Cgeq+0%5C%5C%5C%5C+3%2A%28x-%5Cfrac%7B5%7D%7B6%7D%29%5E2-%5Cfrac%7B25%7D%7B12%7D%2B%5Cfrac%7B36%7D%7B12%7D+%5Cgeq+0%5C%5C%5C%5C+3%2A%28x-%5Cfrac%7B5%7D%7B6%7D%29%5E2%2B%5Cfrac%7B11%7D%7B12%7D+%5Cgeq+0%5C%5C%5C%5C+x%5Cin%28-%5Cinfty%3B%5C+%2B%5Cinfty%29 "3*(x^2-\frac{5}{3}x)+3 \geq 0\\\\ 3*(x^2-2*x*\frac{5}{3*2})+3 \geq 0\\\\ 3*[x^2-2*x*\frac{5}{6}+(\frac{5}{6})^2-(\frac{5}{6})^2]+3 \geq 0\\\\ 3*[(x-\frac{5}{6})-(\frac{5}{6})^2]+3 \geq 0\\\\ 3*(x-\frac{5}{6})^2-3*(\frac{5}{6})^2+3 \geq 0\\\\ 3*(x-\frac{5}{6})^2-3*\frac{25}{36}+3 \geq 0\\\\ 3*(x-\frac{5}{6})^2-\frac{25}{12}+\frac{36}{12} \geq 0\\\\ 3*(x-\frac{5}{6})^2+\frac{11}{12} \geq 0\\\\ x\in(-\infty;\ +\infty)")
