Thay x=2 vào biểu thức

\(x^3+4x^2-3x-18=2^3+4.2^2-3.2-18=8+16-6-18=0\)

Do x=2 cho ta \(x^3+4x^2-3x-18=0\) nên với mọi x lớn hơn hoặc bằng 2 ta luôn thu đc biểu thức lớn hơn hoặc bằng 0

\(x^3+4x^2-3x-18\ge0\)

\(\Leftrightarrow x^3+6x^2+9x-2x^2-12x-18\ge0\)

\(\Leftrightarrow x\left(x^2+6x+9\right)-2\left(x^2+6x+9\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+6x+9\right)\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)^2\ge0\)

Từ \(\left\{{}\begin{matrix}x\ge2\Rightarrow x-2\ge0\\\left(x+3\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)^2\ge0\forall x\ge2\) (Đúng !!)

\(a,5-3x\ge0\)

\(\Rightarrow-3x\ge-5\)

\(\Rightarrow x\le\frac{5}{3}\)

\(b,2-4x\le0\)

\(\Rightarrow-4x\le-2\)

\(\Rightarrow x\ge\frac{2}{4}\)

\(\Rightarrow x\ge\frac{1}{2}\)

\(c,4x-7\ge0\)

\(\Rightarrow4x\ge7\)

\(\Rightarrow x\ge\frac{7}{4}\)

\(ĐK:x\ne\frac{-1}{3}\)

\(PT\Leftrightarrow\left(\frac{4x-3}{3x+1}+2\right)\left(x^2+3x+1-4x-7\right)=0\)

\(\Leftrightarrow\left(\frac{10x-1}{3x+1}\right).\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\)\(x=\frac{1}{10}\)hoặc x=3 hoặc x=-2

Vậy...........

\(-4x^2-3x^2\ge0\)

\(\Leftrightarrow-x^2\left(2x^2+3\right)\ge0\)

Vì \(-x^2\le0\Rightarrow-x^2\left(2x^2+3\right)\ge0\Leftrightarrow2x^2+3\le0\)

\(\Leftrightarrow2x^2\le-3\)

\(\Leftrightarrow x^2\le\frac{-3}{2}\)(vô lí)

Vậy \(x\in\phi\)

\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)

\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)