Cho bieu thuc A= 2 can cua x / can cua x (chi 1 mik x) – 2 + can cua 1 mik x +2/ 3-can cua 1 mik x + (4 nhan can cua 1 mk x – 1) / x-5 nhan can 1 mk x + 6.Rut gon bieu thuc A(giai chi tiet kem dap an, co dieu kien xac dinh)
Cho bieu thuc A= 2 can cua x / can cua x (chi 1 mik x) – 2 + can cua 1 mik x +2/ 3-can cua 1 mik x + (4 nhan can cua 1 mk x – 1) / x-5 nhan can 1 mk x + 6.Rut gon bieu thuc A(giai chi tiet kem dap an, co dieu kien xac dinh)
$$\eqalign{
& A = {{2\sqrt x } \over {\sqrt x – 2}} + {{\sqrt x + 2} \over {3 – \sqrt x }} + {{4\left( {\sqrt x – 1} \right)} \over {x – 5\sqrt x + 6}}\,\,\left( {x \ge 0;\,\,x \ne 4;\,\,x \ne 9} \right) \cr
& A = {{2\sqrt x } \over {\sqrt x – 2}} – {{\sqrt x + 2} \over {\sqrt x – 3}} + {{4\left( {\sqrt x – 1} \right)} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} \cr
& A = {{2\sqrt x \left( {\sqrt x – 3} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) + 4\sqrt x – 4} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} \cr
& A = {{2x – 6\sqrt x – \left( {x – 4} \right) + 4\sqrt x – 4} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} \cr
& A = {{2x – 6\sqrt x – x + 4 + 4\sqrt x – 4} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} \cr
& A = {{x – 2\sqrt x } \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} \cr
& A = {{\sqrt x \left( {\sqrt x – 2} \right)} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x – 3} \right)}} \cr
& A = {{\sqrt x } \over {\sqrt x – 3}} \cr} $$