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\(\left|1-2x\right|< 5-x\)
\(\Leftrightarrow-\left(5-x\right)< 1-2x< 5-x\)
\(\Leftrightarrow x-5< 1-2x< 5-x\)
\(\Leftrightarrow-4< x< 2\)
Bài 3:
c: \(\Leftrightarrow x\left(x-5\right)=0\)
hay x=5
5) \(\sqrt{x+1}+\sqrt{2x+3}=x^2-x-1\) (ĐKXĐ: \(x\ge-1\))
<=>\(\left(\sqrt{x+1}-2\right)+\left(\sqrt{2x+3}-3\right)-\left(x^2-x-6\right)=0\)
<=>\(\dfrac{x-3}{\sqrt{x+1}+2}+\dfrac{2\left(x-3\right)}{\sqrt{2x+3}+3}-\left(x-3\right)\left(x+2\right)=0\)
<=>\(\left(x-3\right)\left(\dfrac{1}{\sqrt{x+1}+2}+\dfrac{2}{\sqrt{2x+3}+3}-x-2\right)=0\)
<=>\(\left[{}\begin{matrix}x-3=0\left(1\right)\\\dfrac{1}{\sqrt{x+1}+2}+\dfrac{2}{\sqrt{2x+3}+3}-x-2=0\left(2\right)\end{matrix}\right.\)
Giải (1) được x=3 thỏa mãn ĐKXĐ.
Giải (2): Từ \(x\ge-1\) ta có:
\(\sqrt{x+1}+2\ge2\), \(\sqrt{2x+3}+3\ge\sqrt{1}+3=4\), \(-x\le1\), từ đó:
VT(2)\(\le\dfrac{1}{2}+\dfrac{2}{4}+1-2=0\).
Như vậy để (2) xảy ra thì x=\(-1\), thỏa mãn ĐKXĐ.
Vậy \(S=\left\{-1;3\right\}\).
Câu 1:
\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)
Câu 2:
\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)
\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)