Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 151241 by bagjagugum123 last updated on 19/Aug/21

Answered by hknkrc46 last updated on 19/Aug/21

$$\left(2\right)\mid 2\mathit{x}-3\mid <\mid \mathit{x}-1\mid$$$$\Rightarrow {(2\mathit{x}-3)}^2<{(\mathit{x}-1)}^2$$$$\Rightarrow 4{\mathit{x}}^2-12\mathit{x}+9<{\mathit{x}}^2-2\mathit{x}+1$$$$\Rightarrow 3{\mathit{x}}^2-10\mathit{x}+8<0$$$$\Rightarrow \left(\underset{=0\Rightarrow \mathit{x}=\frac{4}{3}}{\underbrace{3\mathit{x}-4}}\right)\left(\underset{=0\Rightarrow \mathit{x}=2}{\underbrace{\mathit{x}-2}}\right)<0$$$$\Rightarrow \mathit{x}\in (\frac{4}{3},2)$$$$\left(3\right)\mid \frac{2\mathit{x}-1}{1-3\mathit{x}}\mid >2\Rightarrow \begin{array}{c}\left(\mathit{i}\right)\frac{2\mathit{x}-1}{1-3\mathit{x}}>2\\ \left(\mathit{i}\mathit{i}\right)\frac{2\mathit{x}-1}{1-3\mathit{x}}<-2\end{array}$$$$\left(\mathit{i}\right)\frac{2\mathit{x}-1}{1-3\mathit{x}}>2\Rightarrow \frac{2\mathit{x}-1}{1-3\mathit{x}}-2>0$$$$\Rightarrow \frac{8\mathit{x}-3}{1-3\mathit{x}}>0\to \begin{array}{c}8\mathit{x}-3=0\Rightarrow \mathit{x}=\frac{3}{8}\\ 1-3\mathit{x}=0\Rightarrow \mathit{x}=\frac{1}{3}\end{array}$$$$\Rightarrow \mathit{x}\in (\frac{1}{3},\frac{3}{8})$$$$\left(\mathit{i}\mathit{i}\right)\frac{2\mathit{x}-1}{1-3\mathit{x}}<-2\Rightarrow \frac{2\mathit{x}-1}{1-3\mathit{x}}+2<0$$$$\Rightarrow \frac{-4\mathit{x}+1}{-3\mathit{x}+1}<0\to \begin{array}{c}-4\mathit{x}+1=0\Rightarrow \mathit{x}=\frac{1}{4}\\ -3\mathit{x}+1=0\Rightarrow \mathit{x}=\frac{1}{3}\end{array}$$$$\Rightarrow \mathit{x}\in (\frac{1}{4},\frac{1}{3})$$$$\left(4\right)\mid \frac{2\mathit{x}-1}{1-3\mathit{x}}\mid <1\Rightarrow -1<\frac{2\mathit{x}-1}{1-3\mathit{x}}<1$$$$\Rightarrow \begin{array}{c}\left(\mathit{i}\right)-1<\frac{2\mathit{x}-1}{1-3\mathit{x}}\\ \left(\mathit{i}\mathit{i}\right)\frac{2\mathit{x}-1}{1-3\mathit{x}}<1\end{array}$$$$\left(\mathit{i}\right)-1<\frac{2\mathit{x}-1}{1-3\mathit{x}}\Rightarrow \frac{2\mathit{x}-1}{1-3\mathit{x}}+1>0$$$$\Rightarrow \frac{-\mathit{x}}{1-3\mathit{x}}>0\to \begin{array}{c}-\mathit{x}=0\Rightarrow \mathit{x}=0\\ 1-3\mathit{x}=0\Rightarrow \mathit{x}=\frac{1}{3}\end{array}$$$$\Rightarrow \mathit{x}\in (-\mathrm{\infty },0)\cup (\frac{1}{3},\mathrm{\infty })$$$$\left(\mathit{i}\mathit{i}\right)\frac{2\mathit{x}-1}{1-3\mathit{x}}<1\Rightarrow \frac{2\mathit{x}-1}{1-3\mathit{x}}-1<0$$$$\Rightarrow \frac{5\mathit{x}-2}{1-3\mathit{x}}<0\to \begin{array}{c}5\mathit{x}-2=0\Rightarrow \mathit{x}=\frac{2}{5}\\ 1-3\mathit{x}=0\Rightarrow \mathit{x}=\frac{1}{3}\end{array}$$$$\Rightarrow \mathit{x}\in (-\mathrm{\infty },\frac{1}{3})\cup (\frac{2}{5},\mathrm{\infty })$$$$\left(5\right)\mid \mathit{x}+3\mid <2\mid \mathit{x}-2\mid$$$$\Rightarrow {(\mathit{x}+3)}^2<4{(\mathit{x}-2)}^2$$$$\Rightarrow {\mathit{x}}^2+6\mathit{x}+9<16{\mathit{x}}^2-16\mathit{x}+16$$$$\Rightarrow 15{\mathit{x}}^2-22\mathit{x}+7>0$$$$\Rightarrow \left(\underset{=0\Rightarrow \mathit{x}=\frac{7}{15}}{\underbrace{15\mathit{x}-7}}\right)\left(\underset{=0\Rightarrow \mathit{x}=1}{\underbrace{\mathit{x}-1}}\right)>0$$$$\Rightarrow \mathit{x}\in (-\mathrm{\infty },\frac{7}{15})\cup (1,\mathrm{\infty })$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com