Vì \(\sqrt{x}\ge0\Rightarrow0\le\sqrt{x}< 3\)

\(\Rightarrow0\le x< 9\Rightarrow x\in\left\{0;1;2;...;8\right\}\)

\(b,x^2=5\)

\(\Rightarrow\sqrt{x^2}=\pm\sqrt{5}\)

\(\Rightarrow x=\pm\sqrt{5}\)

c, Ta có:\(x\ne0\left(\sqrt{x}\ge0\forall x\right)\)

\(\Rightarrow\sqrt{0}\le\sqrt{x}< \sqrt{2}\)

\(\Rightarrow0\le x< 4\)

\(\Rightarrow x\in\left\{0;1;2;3\right\}\)

\(\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(=6x^2+9x+14x+21-6x^2-33x+10x+55\)

\(=\left(6x^2-6x^2\right)+\left(9x+14x-33x+10x\right)+\left(21+55\right)\)

\(=76\)

\(\left(x+2\right)\left(x-3\right)-\left(x+3\right)\left(x-4\right)\)

\(=x^2-3x+2x-6-x^2+4x-3x+12\)

\(=\left(x^2-x^2\right)-\left(3x-2x-4x+3x\right)-\left(6-12\right)\)

\(=6\)

Ủng hộ nhé ^^

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

\(\Leftrightarrow\)\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)

\(x^2+y^2+z^2\ge xy+yz+zx\) \(\forall x;y;z\in R\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx\ge0\)\(\forall x;y;z\in R\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)\(\forall x;y;z\in R\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)\(\forall x;y;z\in R\)

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

                                                                          đpcm

Tham khảo nhé