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\(y=\dfrac{\left(x-1\right)\left(3-2x\right)}{2x-4}>0\)
nghiệm của y: x - 1 = 0 <=> x = 1
3 - 2x = 0 <=> x = 3/2
y không xác định: 2x - 4 = 0 <=> x = 2
x | -∞ 1 3/2 2 +∞ |
x - 1 | - 0 + | + | + |
3 - 2x | - | - 0 + | + |
2x - 4 | - | - | - 0 + |
dấu y | - 0 + 0 - || + |
vậy: \(S=\left(1;\dfrac{3}{2}\right)\cup\left(2;+\text{∞}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b.
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) hệ tương đương:
\(\left\{{}\begin{matrix}\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\\\dfrac{1}{x^3}+y^3=19\end{matrix}\right.\)
Đặt \(\left(\dfrac{1}{x};y\right)=\left(u;v\right)\) ta được: \(\left\{{}\begin{matrix}uv^2+u^2v=-6\\u^3+v^3=19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3uv^2+3u^2v=-18\\u^3+v^3+19\end{matrix}\right.\)
Cộng vế với vế:
\(\left(u+v\right)^3=1\Rightarrow u+v=1\)
Thay vào \(u^2v+uv^2=-6\Rightarrow uv=-6\)
Theo Viet đảo, u và v là nghiệm của:
\(t^2-t-6=0\) \(\Rightarrow\left[{}\begin{matrix}t=-2\\t=3\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(-2;3\right);\left(3;-2\right)\)
\(\Rightarrow\left(\dfrac{1}{x};y\right)=\left(-2;3\right);\left(3;-2\right)\)
\(\Rightarrow\left(x;y\right)=\left(-\dfrac{1}{2};3\right);\left(\dfrac{1}{3};-2\right)\)
a.
ĐKXĐ: \(x\ne3\)
- Với \(x\ge0\) pt trở thành:
\(\dfrac{x^2-x-12}{x-3}=2x\Rightarrow x^2-x-12=2x^2-6x\)
\(\Leftrightarrow x^2-5x+12=0\) (vô nghiệm)
- Với \(x< 0\) pt trở thành:
\(\dfrac{x^2+x-12}{x-3}=2x\Rightarrow\dfrac{\left(x-3\right)\left(x+4\right)}{x-3}=2x\)
\(\Rightarrow x+4=2x\Rightarrow x=4>0\) (ktm)
Vậy pt đã cho vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
b/
\(\frac{3x+5}{2x^2-5x+3}\geq 0\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} 3x+5\geq 0\\ 2x^2-5x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x+5\leq 0\\ 2x^2-5x+3<0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq \frac{-5}{3}\\ x>\frac{3}{2}(\text{hoặc}) x< 1\end{matrix}\right.\\ \left\{\begin{matrix} x\leq \frac{-5}{3}\\ 1< x< \frac{3}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} x>\frac{3}{2}\\ \frac{-5}{3}\leq x< 1\end{matrix}\right.\ \)
c/
$2x^3+x+3>0$
$\Leftrightarrow 2x^2(x+1)-2x(x+1)+3(x+1)>0$
$\Leftrightarrow (x+1)(2x^2-2x+3)>0$
$\Leftrightarrow (x+1)[x^2+(x-1)^2+2]>0$
$\Leftrightarrow x+1>0$
$\Leftrightarrow x>-1$
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)
TH1 : \(x\le-3\) ( LĐ )
TH2 : \(x\ge0\)
BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)
\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)
\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow x\ge0\)
Vậy \(S=R/\left(-3;0\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x^2-3}{x^2-2x}>=2\)
=>\(\dfrac{x^2-3-2x^2+4x}{x^2-2x}>=0\)
=>\(\dfrac{-x^2+4x-3}{x^2-2x}>=0\)
=>\(\dfrac{x^2-4x+3}{x^2-2x}< =0\)
Theo BXD, ta có: 0<x<=1 hoặc 2<x<=3
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\(\Leftrightarrow\left(\sqrt[3]{x+1}-1\right)+\left(\sqrt{2x+4}-2\right)< -x\sqrt{2}\)
=>\(\dfrac{x+1-1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{2x+4-4}{\sqrt{2x+4}+2}+x\sqrt{2}< 0\)
=>x<0
=>-1<x<0
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(2x – 1)(x + 3) – 3x + 1 ≤ (x – 1)(x + 3) + x2 – 5
⇔ 2x2 + 6x - x – 3 – 3x + 1 ≤ x2 + 3x - x – 3 + x2 – 5
⇔ 2x2 + 2x – 2 ≤ 2x2 + 2x – 8
⇔ 6 ≤ 0 (Vô lý).
Vậy BPT vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x^2-3x-3}{3-2x}\ge1\\ \Leftrightarrow x^2-3x-3\ge3-2x\\ \Leftrightarrow x^2-3x+2x-3-3\ge0\\ \Leftrightarrow x^2-x-6\ge0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\x-3\ge3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge3\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\x-3\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\le3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)
\(\left|x^2+2x-3\right|< x^2+3\\ \Leftrightarrow\left\{{}\begin{matrix}-x^2-3< x^2+2x-3\\x^2+2x-3< x^2+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x^2+2x>0\\2x+6< 0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< -1\\x>0\end{matrix}\right.\\x< -3\end{matrix}\right.\\ \Leftrightarrow x< -3\)