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a, x2- 2x +8 >0 =>(x-1)2+7>0(dung voi moi x)
=> \(x\in R\)
b, x2- 3x -10 <0 \(\Leftrightarrow x^2-5x+2x-10< 0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)< 0\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2< x< 5\\x>5\end{matrix}\right.\)
c,\(2x^2-3x+4>0\Leftrightarrow2\left(2x^2-3x+4\right)>0\)
\(\Leftrightarrow4x^2-6x+8>0\Leftrightarrow\left(2x-\dfrac{3}{2}\right)^2+\dfrac{23}{4}>0\)
(la dang thuc dung voi moi x)\(\Rightarrow x\in R\)
d, \(6x^2-13x+6\le0\)
\(\Leftrightarrow6x^2-9x-4x+6\le0\Leftrightarrow\left(2x-3\right)\left(3x-2\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\x\ge\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x\le\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}\le x\le\dfrac{3}{2}\\x\ge\dfrac{3}{2}\end{matrix}\right.\)
\(a,2x-6< 0\Leftrightarrow2x>6\Leftrightarrow x>3\)
\(b,5x+2x< 4+25\Leftrightarrow7x< 29\Leftrightarrow x< \frac{29}{7}\)
\(c,-5x+6>8-10+8x\Leftrightarrow-5x-8x>8-10-6\)
\(-13x>-8\Leftrightarrow x< \frac{8}{13}\)
\(d,3x-12\le2-4x\Leftrightarrow3x+4x\le2+12\)
\(\Leftrightarrow7x\le14\Leftrightarrow x\le2\)
\(e,\frac{3\left(x-3\right)}{6}>\frac{2\left(2x-5\right)}{6}+\frac{6}{6}\Rightarrow3x-9>4x-10+6\)
\(\Leftrightarrow3x-4x>-4+9\Leftrightarrow x>-5\)
\(f,3\left(2x-3\right)>1+2\left(2+2x\right)\Leftrightarrow6x-9>1+4+4x\)
\(6x-4x>14\Leftrightarrow2x>14\Leftrightarrow x>7\)
Tự biểu diễn nha!
