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

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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} $$
   $$\eqalign{
   & d)\,\,{S_{ABC}} = {1 \over 2}d\left( {A;BC} \right).BC \cr
   & PT\,\,BC:\,\,y = – {{{{\left( {m + 1} \right)}^2}} \over 4} + m \cr
   & \Rightarrow d\left( {A;BC} \right) = \left| {m + {{{{\left( {m + 1} \right)}^2}} \over 4} – m} \right| = {{{{\left( {m + 1} \right)}^2}} \over 4} \cr
   & BC = 2\sqrt {{{m + 1} \over 2}} \cr
   & \Rightarrow {S_{ABC}} = {1 \over 2}.{{{{\left( {m + 1} \right)}^2}} \over 4}.2\sqrt {{{m + 1} \over 2}} = 8 \cr
   & \Leftrightarrow {\sqrt {{{m + 1} \over 2}} ^5} = 64 \cr
   & \Leftrightarrow \sqrt {{{m + 1} \over 2}} = \root 5 \of {64} \cr
   & \Leftrightarrow {{m + 1} \over 2} = {\left( {\root 5 \of {64} } \right)^2} \Leftrightarrow m + 1 = 2{\left( {\root 5 \of {64} } \right)^2} \cr
   & \Leftrightarrow m = 2{\left( {\root 5 \of {64} } \right)^2} – 1\,\,\left( {tm} \right) \cr
   & e)\,\,Ap\,\,dung\,\,CT:\,\,R = {{abc} \over {4S}} \cr
   & f)\,\,Ap\,\,dung\,\,CT:\,\,r = {S \over p} \cr} $$

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