![-3x^2+5x-3 \leq 0\\\\ 3x^2-5x+3 \geq 0\\\\ ----------------------\\ D=25-4*3*3\ \textless \ 0\\](https://tex.z-dn.net/?f=-3x%5E2%2B5x-3+%5Cleq+0%5C%5C%5C%5C%0A3x%5E2-5x%2B3+%5Cgeq+0%5C%5C%5C%5C%0A----------------------%5C%5C%0AD%3D25-4%2A3%2A3%5C+%5Ctextless+%5C+0%5C%5C)
по скольку дискриминант отрицателен, то левая часть неравенства положительна при любом действительном значении
![x](https://tex.z-dn.net/?f=x)
Альтернатива - выделить полный квадрат:
![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%0A3%2A%28x%5E2-2%2Ax%2A%5Cfrac%7B5%7D%7B3%2A2%7D%29%2B3+%5Cgeq+0%5C%5C%5C%5C%0A3%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%0A3%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%0A3%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%0A3%2A%28x-%5Cfrac%7B5%7D%7B6%7D%29%5E2-3%2A%5Cfrac%7B25%7D%7B36%7D%2B3+%5Cgeq+0%5C%5C%5C%5C%0A3%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%0A3%2A%28x-%5Cfrac%7B5%7D%7B6%7D%29%5E2%2B%5Cfrac%7B11%7D%7B12%7D+%5Cgeq+0%5C%5C%5C%5C%0Ax%5Cin%28-%5Cinfty%3B%5C+%2B%5Cinfty%29)
Ответ:
![(-\infty;\ +\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%3B%5C+%2B%5Cinfty%29)
B^2-6b+9=(b-3)^2. подставляем числовое значение: (123-3)^2=14400.
2x + y = 12
7x - y = 3
------------------ + сложим
9x = 15
x = 5/3 = 1 целая 2/3
y = 12 - 2*5/3 = 12 - 10/3 = 26/3 = 8 целых 2/3