8-3x=√x+2. решите пожалуйста очень надо
справа корень под всем выражением?
да
Одз: теперь возводим обе части в квадрат: Ответ: x=2
![\left \{ {{8-3x \geq 0} \atop {x+2 \geq 0}} \right.
\\ \left \{ {{x \leq \frac{8}{3} } \atop {x \geq -2}} \right.
\\ x \in [-2; \frac{8}{3} ]](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B8-3x+%5Cgeq+0%7D+%5Catop+%7Bx%2B2+%5Cgeq+0%7D%7D+%5Cright.+%0A%5C%5C+%5Cleft+%5C%7B+%7B%7Bx+%5Cleq++%5Cfrac%7B8%7D%7B3%7D+%7D+%5Catop+%7Bx+%5Cgeq+-2%7D%7D+%5Cright.+%0A%5C%5C+x+%5Cin+%5B-2%3B+%5Cfrac%7B8%7D%7B3%7D+%5D "\left \{ {{8-3x \geq 0} \atop {x+2 \geq 0}} \right.
\\ \left \{ {{x \leq \frac{8}{3} } \atop {x \geq -2}} \right.
\\ x \in [-2; \frac{8}{3} ]")
![8-3x=\sqrt{x+2}
\\(8-3x)^2=x+2
\\64-48x+9x^2-x-2=0
\\9x^2-49x+62=0
\\D=49^2-4*9*62=2401-2232=169=13^2
\\x_1= \frac{49+13}{18} = \frac{62}{18} = \frac{31}{9} \notin[-2; \frac{8}{3} ]
\\x_2= \frac{49-13}{18} = \frac{36}{18} =2 \in [-2; \frac{8}{3} ]](https://tex.z-dn.net/?f=8-3x%3D%5Csqrt%7Bx%2B2%7D%0A%5C%5C%288-3x%29%5E2%3Dx%2B2%0A%5C%5C64-48x%2B9x%5E2-x-2%3D0%0A%5C%5C9x%5E2-49x%2B62%3D0%0A%5C%5CD%3D49%5E2-4%2A9%2A62%3D2401-2232%3D169%3D13%5E2%0A%5C%5Cx_1%3D+%5Cfrac%7B49%2B13%7D%7B18%7D+%3D+%5Cfrac%7B62%7D%7B18%7D+%3D+%5Cfrac%7B31%7D%7B9%7D+%5Cnotin%5B-2%3B+%5Cfrac%7B8%7D%7B3%7D+%5D%0A%5C%5Cx_2%3D+%5Cfrac%7B49-13%7D%7B18%7D+%3D+%5Cfrac%7B36%7D%7B18%7D+%3D2+%5Cin+%5B-2%3B+%5Cfrac%7B8%7D%7B3%7D+%5D "8-3x=\sqrt{x+2}
\\(8-3x)^2=x+2
\\64-48x+9x^2-x-2=0
\\9x^2-49x+62=0
\\D=49^2-4*9*62=2401-2232=169=13^2
\\x_1= \frac{49+13}{18} = \frac{62}{18} = \frac{31}{9} \notin[-2; \frac{8}{3} ]
\\x_2= \frac{49-13}{18} = \frac{36}{18} =2 \in [-2; \frac{8}{3} ]")