Giải các bất phương trình: x - 1 2 < x(x + 3)
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1) Ta có: \(4x+8=3x-1\)
\(\Leftrightarrow4x-3x=-1-8\)
\(\Leftrightarrow x=-9\)
2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)
\(\Leftrightarrow10-5x-15-3x+3>0\)
\(\Leftrightarrow-8x>2\)
hay \(x< \dfrac{-1}{4}\)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
1:
ĐKXĐ: x<>3
\(\dfrac{x-1}{x-3}>1\)
=>\(\dfrac{x-1-\left(x-3\right)}{x-3}>0\)
=>\(\dfrac{x-1-x+3}{x-3}>0\)
=>\(\dfrac{2}{x-3}>0\)
=>x-3>0
=>x>3
2: ĐKXĐ: \(\left[{}\begin{matrix}x>=3\\x< =-4\end{matrix}\right.\)
\(\sqrt{x^2+x-12}< 8-x\)
=>\(\left\{{}\begin{matrix}8-x>=0\\x^2+x-12< \left(8-x\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =8\\x^2+x-12-x^2+16x-64< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =8\\17x-76< 0\end{matrix}\right.\)
=>\(x< \dfrac{76}{17}\)
Kết hợp ĐKXĐ, ta được: \(\left[{}\begin{matrix}3< =x< \dfrac{76}{17}\\x< =-4\end{matrix}\right.\)
Ta có: 3(x – 2)(x + 2) < 3 x 2 + x
⇔ 3( x 2 – 4) < 3 x 2 + x
⇔ 3 x 2 – 12 < 3 x 2 + x
⇔ 3 x 2 – 3 x 2 – x < 12
⇔ -x < 12
⇔ x > -12
Vậy tập nghiệm của bất phương trình là S = {x|x > -12}
1) \(ĐK:x\ne2\)
Nếu \(x>2\)
BPT ⇔ \(x^2-2x+5-\left(x-1\right)\left(x-2\right)\ge0\) ⇔ \(x^2-2x+5-\left(x^2-3x+3\right)\ge0\)
⇔\(x+2\ge0\) ⇔\(x\ge-2\) ⇒ Lấy \(x\ge2\)
Nếu \(x< 2\)
BPT ⇔\(\dfrac{-\left(x^2-2x+5\right)}{x-2}-x+1\ge0\) ⇔\(-x^2+2x-5-\left(x-1\right)\left(x-2\right)\ge0\)
⇔\(-x^2+2x-5-x^2+3x-2\ge0\)
⇔\(-2x^2+5x-7\ge0\)
⇔\(x^2-\dfrac{5}{2}x+\dfrac{7}{2}\le0\)
⇔\(\left(x-\dfrac{5}{4}\right)^2\le\dfrac{11}{4}\)
⇔\(\left[{}\begin{matrix}x-\dfrac{5}{4}\le\dfrac{11}{4}\\x-\dfrac{5}{4}\le\dfrac{-11}{4}\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x\le4\\x\le\dfrac{-3}{2}\end{matrix}\right.\) ⇔ \(x\le\dfrac{-3}{2}\)
S= [2;+∞)U(-∞;\(\dfrac{-3}{2}\)]
2) \(ĐK:x\ne-1\)
Nếu \(x>-1\)
BPT ⇔ \(2x-3-2\left(x+1\right)< 0\) ⇔\(2x-3-2x-2< 0\)
⇔\(-5< 0\) ( luôn đúng với mọi \(x>-1\))
Nếu \(x< -1\)
BPT⇔\(\dfrac{-\left(2x-3\right)}{x+1}-2< 0\) ⇔\(-\left(2x-3\right)-2\left(x+1\right)< 0\) ⇔\(-4x+1< 0\) ⇔ \(x>\dfrac{-1}{4}\)
Vậy S=....
1.
$2(-2x+1)\leq -x+3$
$\Leftrightarrow -4x+2\leq -x+3$
$\Leftrightarrow -1\leq 3x$
$\Leftrightarrow x\geq \frac{-1}{3}$
2.
$2(x+1)\leq -x+3$
$\Leftrightarrow 2x+2\leq -x+3$
$\Leftrightarrow 3x\leq 1$
$\Leftrightarrow x\leq \frac{1}{3}$
3.
$5-3(x-1)>2$
$\Leftrightarrow 5-(3x-3)>2$
$\Leftrightarrow 8-3x>2$
$\Leftrightarrow 8-3x-2>0$
$\Leftrightarrow 6-3x>0$
$\Leftrightarrow 6>3x$
$\Leftrightarrow x< 2$
4.
$x^2-12x+3-(x-3)^2>0$
$\Leftrightarrow x^2-12x+3-(x^2-6x+9)>0$
$\Leftrightarrow -6x-6>0$
$\Leftrightarrow -6>6x$
$\Leftrightarrow x< -1$
Ta có: x - 1 2 < x(x + 3) ⇔ x 2 – 2x + 1 < x 2 + 3x
⇔ x 2 – 2x + 1 – x 2 – 3x < 0
⇔ -5x + 1 < 0 ⇔ 5x > 1 ⇔ x > 1/5.
Vậy tập nghiệm của bất phương trình là: {x| x > 1/5}