Question Number 211598 by MrGaster last updated on 14/Sep/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{cases}{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{3}<\mathrm{0}}\\{\frac{\mathrm{2}\boldsymbol{{x}}−\mathrm{1}}{\boldsymbol{{x}}+\mathrm{2}}\geq\mathrm{1}}\end{cases} \\ $$$$ \\ $$
Answered by A5T last updated on 14/Sep/24
$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}=\left({x}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)<\mathrm{0}\Rightarrow\left(\mathrm{1},\mathrm{3}\right)={A} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{2}}\geqslant\mathrm{1}\Leftrightarrow\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{2}}−\mathrm{1}\geqslant\mathrm{0}\Leftrightarrow\frac{{x}−\mathrm{3}}{{x}+\mathrm{2}}\geqslant\mathrm{0} \\ $$$$\Rightarrow\left(−\infty,−\mathrm{2}\right)\:\cup\:\left[\mathrm{3},+\infty\right)={B} \\ $$$${A}\cap{B}=\emptyset\Rightarrow\:{No}\:{solution}\:{exists} \\ $$
Answered by Faetmaaa last updated on 15/Sep/24
![x^2 −4x+3 = (x−2)^2 −1 x^2 −4x+3 < 0 (x−2)^2 −1 < 0 ∣x−2∣ < 1 x∈(1, 3) ((2x−1)/(x+2)) ≤ 1 2x−1 ≤ x+2 x ≤ 3 x∈(−∞, 3]\{2} x∈(1, 3)\{2}](https://www.tinkutara.com/question/Q211648.png)
$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\:=\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}\:<\:\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\:<\:\mathrm{0} \\ $$$$\mid{x}−\mathrm{2}\mid\:<\:\mathrm{1} \\ $$$${x}\in\left(\mathrm{1},\:\mathrm{3}\right) \\ $$$$ \\ $$$$\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{2}}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{2}{x}−\mathrm{1}\:\leqslant\:{x}+\mathrm{2} \\ $$$${x}\:\leqslant\:\mathrm{3} \\ $$$${x}\in\left(−\infty,\:\mathrm{3}\right]\backslash\left\{\mathrm{2}\right\} \\ $$$$ \\ $$$$\boldsymbol{{x}}\in\left(\mathrm{1},\:\mathrm{3}\right)\backslash\left\{\mathrm{2}\right\} \\ $$
Commented by Frix last updated on 15/Sep/24
$$\mathrm{The}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{condition}\:\mathrm{is}\:\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{2}}\geqslant\mathrm{1} \\ $$