Tìm m để đths y=x^4 - (m+1)x^2 +m sao cho : a, Tam giác ABC vuông cân b, Tam giác ABC đều c, Tam giác ABC coa 1 góc = 120 độ d, Diện tích tam giác ABC=8 e, Bán kính đt ngoại tiếp=1 f, Bán kính đt nội tiếp =1

$$\eqalign{ & y = {x^4} - \left( {m + 1} \right){x^2} + m \cr & y' = 4{x^3} - 2\left( {m + 1} \right)x = 0 \Leftrightarrow 2x\left[ {2{x^2} - \left( {m + 1} \right)} \right] = 0 \cr & \Leftrightarrow \left[ \matrix{ x = 0 \Rightarrow y = m \hfill \cr {x^2} = {{m + 1} \over 2}\,\,\left( 1 \right) \hfill \cr} \right. \cr & De\,\,ham\,\,so\,\,co\,\,3\,\,CT\,\, \Rightarrow \left( 1 \right)\,\,co\,\,2\,\,nghiem\,\,pb \ne 0 \cr & \Rightarrow {{m + 1} \over 2} > 0 \Leftrightarrow m > - 1 \cr & \left( 1 \right) \Leftrightarrow x = \pm \sqrt {{{m + 1} \over 2}} \cr & \Rightarrow y = {{{{\left( {m + 1} \right)}^2}} \over 4} - \left( {m + 1} \right){{m + 1} \over 2} + m \cr & y = - {{{{\left( {m + 1} \right)}^2}} \over 4} + m \cr & \Rightarrow A\left( {0;m} \right);\,\,B\left( {\sqrt {{{m + 1} \over 2}} ; - {{{{\left( {m + 1} \right)}^2}} \over 4} + m} \right);\,\,C\left( { - \sqrt {{{m + 1} \over 2}} ; - {{{{\left( {m + 1} \right)}^2}} \over 4} + m} \right) \cr & \Rightarrow \Delta ABC\,\,can\,\,tai\,\,A. \cr & a)\,\,\Delta ABC\,\,vuong\,\,can\,\,tai\,\,A \cr & \Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 0 \cr & \overrightarrow {AB} = \left( {\sqrt {{{m + 1} \over 2}} ; - {{{{\left( {m + 1} \right)}^2}} \over 4}} \right);\,\,\overrightarrow {AC} = \left( { - \sqrt {{{m + 1} \over 2}} ; - {{{{\left( {m + 1} \right)}^2}} \over 4}} \right) \cr & \overrightarrow {AB} .\overrightarrow {AC} = 0 \cr & \Leftrightarrow - {{m + 1} \over 2} + {{{{\left( {m + 1} \right)}^4}} \over {16}} = 0 \cr & \Leftrightarrow {\left( {m + 1} \right)^4} - 8\left( {m + 1} \right) = 0 \cr & \Leftrightarrow \left( {m + 1} \right)\left[ {{{\left( {m + 1} \right)}^3} - 8} \right] = 0 \cr & \Leftrightarrow \left[ \matrix{ m + 1 = 0 \hfill \cr m + 1 = 2 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ m = - 1\,\,\left( {loai} \right) \hfill \cr m = 1\,\,\,\left( {tm} \right) \hfill \cr} \right. \cr & Vay\,\,m = 1. \cr & b)\,\,\Delta ABC\,\,deu\,\, \Leftrightarrow A{B^2} = B{C^2} \cr & \Leftrightarrow {{m + 1} \over 2} + {{{{\left( {m + 1} \right)}^4}} \over {16}} = 4.{{m + 1} \over 2} \cr & \Leftrightarrow {{{{\left( {m + 1} \right)}^4}} \over {16}} = 3{{m + 1} \over 2} \cr & \Leftrightarrow {\left( {m + 1} \right)^4} = 24\left( {m + 1} \right) \cr & \Leftrightarrow \left( {m + 1} \right)\left[ {{{\left( {m + 1} \right)}^3} - 24} \right] = 0 \cr & \Leftrightarrow \left[ \matrix{ m = - 1\,\,\left( {loai} \right) \hfill \cr m + 1 = \root 3 \of {24} \hfill \cr} \right. \Leftrightarrow m = \root 3 \of {24} - 1\,\,\left( {tm} \right) \cr & c)\,\,\,\Delta ABC\,\,co\,\,1\,\,goc\,\,{120^0} \Rightarrow \widehat A = {120^0} \cr & \Rightarrow \cos A = {{\overrightarrow {AB} .\overrightarrow {AC} } \over {AB.AC}} = {{\left( {m + 1} \right)\left[ {{{\left( {m + 1} \right)}^3} - 8} \right]} \over {{{m + 1} \over 2} + {{{{\left( {m + 1} \right)}^4}} \over {16}}}} = - {1 \over 2} \cr & \Leftrightarrow - \left( {m + 1} \right)\left[ {{{\left( {m + 1} \right)}^3} - 8} \right] = {{m + 1} \over 2} + {{{{\left( {m + 1} \right)}^4}} \over {16}} \cr & \Leftrightarrow - \left[ {{{\left( {m + 1} \right)}^3} - 8} \right] = {1 \over 2} + {{{{\left( {m + 1} \right)}^3}} \over {16}}\,\,\left( {Do\,\,m \ne - 1} \right) \cr & \Leftrightarrow - 16\left[ {{{\left( {m + 1} \right)}^3} - 8} \right] = 8 + {\left( {m + 1} \right)^3} \cr & \Leftrightarrow 17{\left( {m + 1} \right)^3} = 120 \cr & \Leftrightarrow {\left( {m + 1} \right)^3} = {{120} \over {17}} \cr & \Leftrightarrow m + 1 = \root 3 \of {{{120} \over {17}}} \cr & \Leftrightarrow m = \root 3 \of {{{120} \over {17}}} - 1\,\,\left( {tm} \right) \cr} $$ $$\eqalign{ & d)\,\,{S_{ABC}} = {1 \over 2}d\left( {A;BC} \right).BC \cr & PT\,\,BC:\,\,y = - {{{{\left( {m + 1} \right)}^2}} \over 4} + m \cr & \Rightarrow d\left( {A;BC} \right) = \left| {m + {{{{\left( {m + 1} \right)}^2}} \over 4} - m} \right| = {{{{\left( {m + 1} \right)}^2}} \over 4} \cr & BC = 2\sqrt {{{m + 1} \over 2}} \cr & \Rightarrow {S_{ABC}} = {1 \over 2}.{{{{\left( {m + 1} \right)}^2}} \over 4}.2\sqrt {{{m + 1} \over 2}} = 8 \cr & \Leftrightarrow {\sqrt {{{m + 1} \over 2}} ^5} = 64 \cr & \Leftrightarrow \sqrt {{{m + 1} \over 2}} = \root 5 \of {64} \cr & \Leftrightarrow {{m + 1} \over 2} = {\left( {\root 5 \of {64} } \right)^2} \Leftrightarrow m + 1 = 2{\left( {\root 5 \of {64} } \right)^2} \cr & \Leftrightarrow m = 2{\left( {\root 5 \of {64} } \right)^2} - 1\,\,\left( {tm} \right) \cr & e)\,\,Ap\,\,dung\,\,CT:\,\,R = {{abc} \over {4S}} \cr & f)\,\,Ap\,\,dung\,\,CT:\,\,r = {S \over p} \cr} $$