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1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
\(a,\left(x+3\right)^2-3\left(2x-1\right)>x\left(x-4\right)\)
\(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
\(\Leftrightarrow x^2-x^2+4x+6x-6x>-3-9\)
\(\Leftrightarrow4x>-12\)
\(\Leftrightarrow x>-3\)
\(d,\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1}{x-3}-2\le0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)}{x-3}-\dfrac{2\left(x-3\right)}{x-3}\le0\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}\le0\)
\(\Leftrightarrow\dfrac{7}{x-3}\le0\)
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy BPT có nghiệm x ≥ 3
\(e,\dfrac{12-3x}{2x+6}>3\)
\(\Leftrightarrow\dfrac{12-3x}{2x+6}-3>0\)
\(\Leftrightarrow\dfrac{12-3x}{2x+6}-\dfrac{3\left(2x+6\right)}{2x+6}>0\)
\(\Leftrightarrow\dfrac{12-3x-6x-18}{2x+6}>0\)
\(\Leftrightarrow\dfrac{-9x-6}{2x+6}>0\)
\(\Leftrightarrow\dfrac{-3\left(3x+2\right)}{2\left(x+3\right)}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x+2>0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}3x+2< 0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-\dfrac{2}{3}\\x>-3\end{matrix}\right.\\\left\{{}\begin{matrix}x< -\dfrac{2}{3}\\x< -3\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{-3;-\dfrac{2}{3}\right\}\)
\(a,2x-6< 0\Leftrightarrow2x>6\Leftrightarrow x>3\)
\(b,5x+2x< 4+25\Leftrightarrow7x< 29\Leftrightarrow x< \frac{29}{7}\)
\(c,-5x+6>8-10+8x\Leftrightarrow-5x-8x>8-10-6\)
\(-13x>-8\Leftrightarrow x< \frac{8}{13}\)
\(d,3x-12\le2-4x\Leftrightarrow3x+4x\le2+12\)
\(\Leftrightarrow7x\le14\Leftrightarrow x\le2\)
\(e,\frac{3\left(x-3\right)}{6}>\frac{2\left(2x-5\right)}{6}+\frac{6}{6}\Rightarrow3x-9>4x-10+6\)
\(\Leftrightarrow3x-4x>-4+9\Leftrightarrow x>-5\)
\(f,3\left(2x-3\right)>1+2\left(2+2x\right)\Leftrightarrow6x-9>1+4+4x\)
\(6x-4x>14\Leftrightarrow2x>14\Leftrightarrow x>7\)
Tự biểu diễn nha!
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Chọn A.
Kết hợp với điều kiện x là số tự nhiên nhỏ hơn 6 ⇒ x ∈ {3; 4; 5}