x^2>=0

=>\(x\in R\)

Vậy: S=R

Xét \(5-x=0\Leftrightarrow x=5\)

\(x-1=0\Leftrightarrow x=1\)

\(2+3x=0\Leftrightarrow x=-\dfrac{2}{3}\)

Bảng xét dấu:

Để VT\(\le\)0 <=>\(\left[{}\begin{matrix}-\dfrac{2}{3}\le x\le1\\x\ge5\end{matrix}\right.\)

Vậy...

(5-x)(x-1)(2+3x) ≤ 0

↔ 5-x≤0 <=> x≥5 (1)

x-1 ≤ 0<=> x≤ 1 (2)

2+3x ≤ 0 <=> x≤ -2/3 (3)

Từ (1),(2),(3) ta có:

  x≥5 or x≤1 or x≤ -2/3

chúc bạn học tốt !!!

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow2+x+1\le\dfrac{12}{4}-\dfrac{x-1}{4}\)

\(\Leftrightarrow x+3\le\dfrac{13-x}{4}\)

\(\Leftrightarrow\dfrac{4x+12}{4}\le\dfrac{13-x}{4}\)

\(\Leftrightarrow4x+12\le13-x\)

\(\Leftrightarrow4x+x\le13-12\)

\(\Leftrightarrow5x\le1\)

\(\Leftrightarrow x\le\dfrac{1}{5}\)

Vậy: \(x\le\dfrac{1}{5}\) 

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow\dfrac{12x+36}{12}\le\dfrac{33-3x}{12}\)

\(\Leftrightarrow12x+36\le33-3x\)

\(\Leftrightarrow12x+3x\le-36+33\)

\(\Leftrightarrow15x\le-3\)

\(\Leftrightarrow x\le\dfrac{-1}{5}\)

Điều kiện xác định : \(x+2\ne0\) hay \(x\ne-2\)

Ta có : 

\(\frac{x-1}{x+2}< 0\)

Trường hợp 1 : 

\(\hept{\begin{cases}x-1< 0\\x+2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x>-2\end{cases}}}\)

\(\Rightarrow\)\(-2< x< 1\)

Trường hợp 2 : 

\(\hept{\begin{cases}x-1>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>1\\x< -2\end{cases}}}\) ( loại ) 

Vậy \(-2< x< 1\)

Chúc bạn học tốt ~ 

Ta có: \(\dfrac{x-2}{x-1}< =0\)

nên x-1>0 và x-2<=0

=>1<x<=2

\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)

\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)

\(\Leftrightarrow4x-2< 3x+18\)

\(\Leftrightarrow4x-3x< 2+18\)

\(\Leftrightarrow x< 20\)

\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)

\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)

\(\Leftrightarrow5x-11>4x+4\)

\(\Leftrightarrow5x-4x>11+4\)

\(\Leftrightarrow x>15\)

Biến đổi tương đương:

\(3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2xy+2yz+2zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(x=y=z\)

ĐKXĐ: \(x>0\)

\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)

\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)

\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)

\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)

\(\Rightarrow a-3>0\Rightarrow a>3\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)

\(\Leftrightarrow2x-6\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)