$$\eqalign{
& 10)\,\,a)\,\,C = {{{x^2} - \sqrt x } \over {x + \sqrt x + 1}} - {{2x + \sqrt x } \over {\sqrt x }} + {{2\left( {x - 1} \right)} \over {\sqrt x - 1}}\,\,\left( {x > 0;\,\,x \ne 1} \right) \cr
& C = {{\sqrt x \left( {x\sqrt x - 1} \right)} \over {x + \sqrt x + 1}} - {{\sqrt x \left( {2\sqrt x + 1} \right)} \over {\sqrt x }} + {{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)} \over {\sqrt x - 1}} \cr
& C = {{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)} \over {x + \sqrt x + 1}} - 2\sqrt x - 1 + 2\sqrt x + 2 \cr
& C = \sqrt x \left( {\sqrt x - 1} \right) + 1 = x - \sqrt x + 1 \cr
& b)\,\,C = {\left( {\sqrt x } \right)^2} - 2.\sqrt x .{1 \over 2} + {1 \over 4} + {3 \over 4} \cr
& C = {\left( {\sqrt x - {1 \over 2}} \right)^2} + {3 \over 4} \cr
& Ta\,\,co:\,\,\sqrt x \ge 0 \Leftrightarrow \sqrt x - {1 \over 2} \ge - {1 \over 2} \cr
& \Leftrightarrow {\left( {\sqrt x - {1 \over 2}} \right)^2} \ge 0 \Leftrightarrow {\left( {\sqrt x - {1 \over 2}} \right)^2} + {3 \over 4} \ge {3 \over 4} \cr
& Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x - {1 \over 2} = 0 \Leftrightarrow x = {1 \over 4}\,\,\left( {tm} \right) \cr} $$
$$\eqalign{
& a)\,\,B = {{2\sqrt x - 9} \over {x - 5\sqrt x + 6}} - {{\sqrt x + 3} \over {\sqrt x - 2}} - {{2\sqrt x + 1} \over {3 - \sqrt x }}\,\,\left( {x \ge 0;\,\,x \ne 4;\,\,x \ne 9} \right) \cr
& B = {{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr
& B = {{2\sqrt x - 9 - x + 9 + 2x - 4\sqrt x + \sqrt x - 2} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr
& B = {{x - \sqrt x - 2} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} = {{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)} \over {\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} \cr
& B = {{\sqrt x + 1} \over {\sqrt x - 3}} \cr
& b)\,\,B = {{\sqrt x - 3 + 4} \over {\sqrt x - 3}} = 1 + {4 \over {\sqrt x - 3}} \cr
& B \in Z \Rightarrow {4 \over {\sqrt x - 3}} \in Z \Rightarrow \sqrt x - 3 \in U\left( 4 \right) = \left\{ { \pm 1; \pm 2; \pm 4} \right\} \cr
& + )\,\,\sqrt x - 3 = 1 \Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16\,\left( {tm} \right) \cr
& + )\,\,\sqrt x - 3 = - 1 \Leftrightarrow \sqrt x = 2 \Leftrightarrow x = 4\,\,\left( {ktm} \right) \cr
& + )\,\,\sqrt x - 3 = 2 \Leftrightarrow \sqrt x = 5 \Leftrightarrow x = 25\,\,\left( {tm} \right) \cr
& + )\,\,\sqrt x - 3 = - 2 \Leftrightarrow \sqrt x = 1 \Leftrightarrow x = 1\,\,\left( {tm} \right) \cr
& + )\,\,\sqrt x - 3 = 4 \Leftrightarrow \sqrt x = 7 \Leftrightarrow x = 49\,\,\left( {tm} \right) \cr
& + )\,\,\sqrt x - 3 = - 4 \Leftrightarrow \sqrt x = - 1\,\,\left( {loai} \right) \cr
& Vay\,\,x \in \{ 16;25;1;49\} \cr
& c)\,\,B < 1 \Leftrightarrow {{\sqrt x + 1} \over {\sqrt x - 3}} < 1 \cr
& \Leftrightarrow {{\sqrt x + 1} \over {\sqrt x - 3}} - 1 < 0 \Leftrightarrow {{\sqrt x + 1 - \sqrt x + 3} \over {\sqrt x - 3}} < 0 \cr
& \Leftrightarrow {4 \over {\sqrt x - 3}} < 0 \Leftrightarrow \sqrt x - 3 < 0 \Leftrightarrow \sqrt x < 3 \Leftrightarrow x < 9 \cr
& Ket\,\,hop\,\,DK \Rightarrow 0 \le x < 9;\,\,x \ne 4 \cr} $$
