\(\left(2x+1\right)^2+\left(1-x\right)3x\le\left(x+2\right)^2\)

\(\Leftrightarrow4x^2+4x+1+3x-3x^2\le x^2+4x+4\)

\(\Leftrightarrow4x^2+4x+3x-3x^2-x^2-4x\le4-1\)

\(\Leftrightarrow3x\le3\Leftrightarrow x\le1\) vậy \(x\le1\)

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; +∞)

\(3x-\frac{x+2}{3}\le\frac{3\left(x-2\right)}{2}+5-x\)

\(\Leftrightarrow\frac{18x}{6}-\frac{2\left(x+2\right)}{6}\le\frac{9\left(x-2\right)}{6}+\frac{30}{6}-\frac{6x}{6}\)

\(\Rightarrow18x-2x-4\le9x-18+30-6x\)

\(\Leftrightarrow16x-4\le3x+12\)

\(\Leftrightarrow13x\le16\)

\(\Leftrightarrow x\le\frac{16}{13}\)

Vậy bất phương trình có tập nghiệm là: \(S=\left\{x|x\le\frac{16}{13}\right\}\)

nhân 2 vế với 6

18x - 2x - 4<=9x - 18 + 30 - 6x

16x - 4 <=3x + 12

13x <=16

x<=16/13

Nhân 2 vế với 6

\(\Leftrightarrow18x-2x-4\le9x-18+30-6x\)

\(\Leftrightarrow18x-2x-9x+6x\le-18+30+4\)

\(\Leftrightarrow-13x\le-16\)

\(\Leftrightarrow x\ge\frac{16}{13}\)