2x-x(3x+1)≤15-3x(x+2)

2x-3x²-x≤15-3x² -6x

2x-3x²-x+3x² +6x≤15

7x≤15

x≤15/7

\(\dfrac{3x-2}{4}>\dfrac{3x+3}{6}\)

\(\Leftrightarrow3\left(3x-2\right)>2\left(3x+3\right)\)

\(\Leftrightarrow9x-6>6x+6\)

\(\Leftrightarrow9x-6x>6+6\)

\(\Leftrightarrow3x>12\)

\(\Leftrightarrow x>4\)

Vậy bất phương trình có nghiệm x > 4 .

\(\left(x-3\right)\left(x+1\right)\left(2-3x\right)>0.\)

Vậy \(\left(x-3\right)\left(x+1\right)\left(2-3x\right)>0\) khi \(x\in\left(-\infty;-1\right)\cup\left(\dfrac{2}{3};3\right)\cup\left(3;+\infty\right).\)

1) ĐKXĐ: \(\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

ta có: (-6).\(\sqrt{6x^2-18x+12}\) > \(6x^2-18x-60\)

⇔ \(6x^2-18x+12\) + \(2.3.\sqrt{6x^2-18x+12}+9-81\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+3\right)^2-9^2\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+12\right).\left(\sqrt{6x^2-18x+12}-6\right)\) > 0

⇔ \(\sqrt{6x^2-18x+12}-6\) > 0

⇔ \(\sqrt{6x^2-18x+12}>6\)

⇔\(6x^2-18x+12>36\)

⇔ \(6x^2-18x-24>0\)

⇔\(\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\)

đối chiếu ĐKXĐ ban đầu ta được: x ϵ (-∞;-1) \(\cup\)(4;+∞)

b) ĐKXĐ: \(\forall x\) ϵ R

\(\left(x-2\right)\sqrt{x^2+4}-\left(x-2\right)\left(x+2\right)\le0\)

⇔\(\left(x-2\right)\left(\sqrt{x^2+4}-x-2\right)\le0\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\\sqrt{x^2+4}-x-2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\\sqrt{x^2+4}-x-2\ge0\end{matrix}\right.\end{matrix}\right.\)⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x^2+4\le x^2+4x+4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x^2+4\ge x^2+4x+4\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\le0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)

Đối chiếu ĐKXĐ ta được x ϵ ( -∞;0) \(\cup\)( 2; +∞)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-\frac{1}{2}\end{matrix}\right.\)

- Với \(\left[{}\begin{matrix}x=2\\x=-\frac{1}{2}\end{matrix}\right.\) BPT thỏa mãn

- Với \(x\ne\left\{-\frac{1}{2};2\right\}\Rightarrow\sqrt{2x^2-3x-2}>0\) BPT tương đương:

\(x^2-3x\ge0\Rightarrow\left[{}\begin{matrix}x\ge3\\x\le0\end{matrix}\right.\)

Kết hợp lại ta được nghiệm: \(\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge3\\x=2\end{matrix}\right.\)

Đặt \(x^2+3x=a\left(a>=-\dfrac{9}{4}\right)\)

BPT sẽ trở thành \(a>=2+\sqrt{5a+14}\)

=>\(a-2>=\sqrt{5a+14}\)

=>\(\sqrt{5a+14}< =a-2\)

=>\(\left\{{}\begin{matrix}a-2>=0\\5a+14< =\left(a-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\5a+14-a^2+4a-4< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\-a^2+9a+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\a^2-9a-10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left(a-10\right)\left(a+1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left[{}\begin{matrix}a>=10\\a< =-1\end{matrix}\right.\end{matrix}\right.\)

=>a>=10

=>\(x^2+3x>=10\)

=>\(x^2+3x-10>=0\)

=>(x+5)(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =-5\end{matrix}\right.\)

ĐK: \(x\ne\dfrac{1}{2};x\ne-\dfrac{1}{3}\)

\(\dfrac{x+2}{3x+1}\ge\dfrac{x-2}{2x-1}\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(2x-1\right)-\left(x-2\right)\left(3x+1\right)}{\left(3x+1\right)\left(2x-1\right)}\ge0\)

\(\Leftrightarrow\dfrac{2x^2+3x-2-3x^2+5x+2}{6x^2-x-1}\ge0\)

\(\Leftrightarrow\dfrac{-x^2+8x}{6x^2-x-1}\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+8x\ge0\\6x^2-x-1>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}-x^2+8x\le0\\6x^2-x-1< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}0\le x\le8\\\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x\le8\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\\-\dfrac{1}{3}< x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}< x\le0\)

Vậy ...

ĐKXĐ: \(x^2+x-1\ge0\)

\(\Rightarrow3x^2-x+1>3\sqrt{\left(x^2-x+1\right)\left(x^2+x-1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2a^2+b^2>3ab\)

\(\Leftrightarrow\left(2a-b\right)\left(a-b\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}2a< b\\a>b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}< \sqrt{x^2+x-1}\\\sqrt{x^2-x+1}>\sqrt{x^2+x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)< x^2+x-1\\x^2-x+1>x^2+x-1\end{matrix}\right.\)

\(\Leftrightarrow...\) (nhớ kết hợp ĐKXĐ ban đầu)