Х+11,3=54,2:2
х+11,3=27,1
х=27,1-11,3
х=15,8
Найдём область допустимых значений функции и преобразуем выражение.
![y=\frac{x^2-4x+4}{2-x}+\frac{3x^2-4x}{x}\; \; ,\; \; ODZ:\; \left \{ {{2-x\ne 0} \atop {x\ne 0}} \right. \; \left \{ {{x\ne 2} \atop {x\ne 0}} \right. \; \; \Rightarrow \\\\x\in (-\infty ,0)\cup (0,2)\cup (2,+\infty )\\\\\frac{x^2-4x+4}{2-x}+\frac{3x^2-4x}{x}=\frac{(x-2)^2}{-(x-2)}+\frac{x(3x-4)}{x}=-(x-2)+(3x-4)=2x-2\\\\x\ne 0\; \; \to \; \; y\ne -2\\\\x\ne 2\; \; \to \; \; y\ne 2](https://tex.z-dn.net/?f=+y%3D%5Cfrac%7Bx%5E2-4x%2B4%7D%7B2-x%7D%2B%5Cfrac%7B3x%5E2-4x%7D%7Bx%7D%5C%3B+%5C%3B+%2C%5C%3B+%5C%3B+ODZ%3A%5C%3B+%5Cleft+%5C%7B+%7B%7B2-x%5Cne+0%7D+%5Catop+%7Bx%5Cne+0%7D%7D+%5Cright.+%5C%3B+%5Cleft+%5C%7B+%7B%7Bx%5Cne+2%7D+%5Catop+%7Bx%5Cne+0%7D%7D+%5Cright.+%5C%3B+%5C%3B+%5CRightarrow+%5C%5C%5C%5Cx%5Cin+%28-%5Cinfty+%2C0%29%5Ccup+%280%2C2%29%5Ccup+%282%2C%2B%5Cinfty+%29%5C%5C%5C%5C%5Cfrac%7Bx%5E2-4x%2B4%7D%7B2-x%7D%2B%5Cfrac%7B3x%5E2-4x%7D%7Bx%7D%3D%5Cfrac%7B%28x-2%29%5E2%7D%7B-%28x-2%29%7D%2B%5Cfrac%7Bx%283x-4%29%7D%7Bx%7D%3D-%28x-2%29%2B%283x-4%29%3D2x-2%5C%5C%5C%5Cx%5Cne+0%5C%3B+%5C%3B+%5Cto+%5C%3B+%5C%3B+y%5Cne+-2%5C%5C%5C%5Cx%5Cne+2%5C%3B+%5C%3B+%5Cto+%5C%3B+%5C%3B+y%5Cne+2+)
Итак, строим график функции у=2х-2 (прямая) и выкалываем точки (0,-2) и (2,2) .
1) 3:3 3/4+2 2/5*2 1/2-3 5/6=3:15/4+12/5*5/2-23/6=3*4/15+6/1*1/1-23/6=12/15+6-23/6=6 12/15-23/6=102/15-23/6=204/30-115/30=89/30=2 29/30.