a)
\(\left(x-3\right)^2< x^2-5x+4\)
\(\Leftrightarrow x^2-6x+9< x^2-5x+4\)
\(\Leftrightarrow x^2-x^2-6x+5x< 4-9\)
\(\Leftrightarrow-x>-5\)
\(\Leftrightarrow x>5\)
Vây...
b)
\(\left(x-3\right)\left(x+3\right)\le\left(x+2\right)^2+3\)
\(\Leftrightarrow x^2-9\le x^2+4x+9\)
\(\Leftrightarrow x^2-x^2-4x\le9+9\)
\(\Leftrightarrow-4x\le18\)
\(\Leftrightarrow x\ge-4,5\)
Vậy....
Bạn tự biểu diễn trên trục số ha!
c)
\(\dfrac{4x-7}{3}>\dfrac{7-x}{5}\)
\(\Leftrightarrow15.\dfrac{4x-5}{3}< 15.\dfrac{7-x}{5}\)
\(\Leftrightarrow5.\left(4x-5\right)< 3.\left(7-x\right)\)
\(\Leftrightarrow24x-25< 21-3x\)
\(\Leftrightarrow20x+3x< 21+25\)
\(\Leftrightarrow23x< 46\)
\(\Leftrightarrow x< 2\)
Vậy...
d)
\(\dfrac{x+2}{x-3}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2< 0\\x-3< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\)
Vậy...
\(a,\left(x-3\right)^2< x^2-5x+4\)
\(\Leftrightarrow x^2-6x+9-x^2+5x-4< 0\)
\(\Leftrightarrow-x+5< 0\)
\(\Leftrightarrow x>5\)
Vậy BPT có nghiệm x > 5
b,\(\left(x-3\right)\left(x+3\right)\le\left(x+2\right)^2+3\)
\(\Leftrightarrow x^2-9\le x^2+4x+4+3\)
\(\Leftrightarrow x^2-x^2-4x-9-4-3\le0\)
\(\Leftrightarrow-4x-16\le0\)
\(\Leftrightarrow x\ge-4\)
Vậy BPT có nghiệm x ≥ -4
\(c,\dfrac{4x-5}{3}>\dfrac{7-x}{5}\)
\(\Leftrightarrow\dfrac{5\left(4x-5\right)}{15}>\dfrac{3\left(7-x\right)}{15}\)
\(\Leftrightarrow20x-25>21-3x\)
\(\Leftrightarrow20x+3x>21+25\)
\(\Leftrightarrow23x>46\)
\(\Leftrightarrow x>2\)
Vậy BPT có nghiệm x > 2
\(d,\dfrac{x+2}{x-3}< 0\)
ĐKXĐ: x ≠ 3
\(\dfrac{x+2}{x-3}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2< 0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2>0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -2\\x>3\end{matrix}\right.\left(vonghiem\right)\\\left\{{}\begin{matrix}x>-2\\x< 3\end{matrix}\right.\Rightarrow-2< x< 3\end{matrix}\right.\)
Vậy BPT có nghiệm -2 < x < 3