Sa se determine m∈R pt xare x=5 este solutia ecuatiei:m²(x-1)=x-3m+2
Ajutor URGENT VA ROG!;multumesn anticipat!

Inlocuim pe x cu 1
obtinem o ecuatie in m
m²(5-1)=5-3m+2
4m²=7-3m
4m²+3m-7=0
4m²-4+3m-3=0
4(m-1)(m+1)+3(m-1)=0
(m-1)(4m+4+3)=0
(m-1) (4m+7)=0
m1=-7/4
m2=1
as simple as that!!
Obs
Nu imi place cuΔ; daca pot, il evit..:::)))
verificare cu Δ
4m²+3m-7=0
m1,2=(-3+-√(9+112))/8=(-3+-11)/8
m1=-14/8=-7/4
m2=8/8=1

Answer Link

Tcostel Tcostel · Answer 2 · 2018-07-06T06:51:14+03:00

[tex]\displaystyle\\ m^2(x-1)=x-3m+2\\ m^2(x-1)-x+3m-2=0\\ m^2(x-1)-x+3m-2=0\\ (x-1)\cdot m^2+3m -x-2=0\\\\ x=5\\\\ (5-1)\cdot m^2+3m -5-2=0\\ 4m^2+3m -7=0\\\\ m_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{-3\pm \sqrt{3^2+4\cdot4\cdot7}}{2\cdot4}=\\\\ =\frac{-3\pm \sqrt{9+112}}{8}=\frac{-3\pm \sqrt{121}}{8}=\frac{-3\pm 11}{8}\\\\ m_1=\frac{-3- 11}{8}=\frac{-14}{8}=\boxed{-\frac{7}{4}}\\\\ m_2=\frac{-3+ 11}{8}=\frac{8}{8}=\boxed{1}[/tex]

[tex]\displaystyle\\ \text{Verificare:}\\ \text{Pentru }~~\boxed{m = 1}~~\text{rezulta:}\\ (x-1)\cdot 1+3 -x-2=0\\ x-x = 2-3+1\\ 0x = 0\\ \text{Ecuatia are o infinitate de solutii printre care: }~\boxed{x=5}\\\\ \text{Pentru }~~\boxed{m=-\frac{7}{4}}~~\text{rezulta:}\\\\ m^2(x-1)-x+3m-2=0\\\\ \Big(-\frac{7}{4}\Big)^2 (x-1)-x+3\cdot\Big(-\frac{7}{4} \Big)-2=0\\\\ \frac{49}{16} (x-1)-x-3\cdot\frac{7}{4}-2=0\\\\ \frac{49}{16}x -\frac{49}{16}-x-\frac{21}{4}-2=0\\\\ [/tex]

[tex]\displaystyle\\ \frac{49}{16}x -x-\frac{49}{16}-\frac{21}{4}-2=0\\\\ \frac{49}{16}x -x =\frac{49}{16}+\frac{21}{4}+2\\\\ \frac{49-16}{16}x =\frac{49}{16}+\frac{84}{16}+\frac{32}{16}\\\\ \frac{33}{16}x =\frac{49+84+32}{16}\\\\ \frac{33}{16}x =\frac{165}{16}~~~\Big| \cdot 16\\\\ 33x=165\\\\ x= \frac{165}{33}=5\\\\ \boxed{x=5} [/tex]