Câu 1: Giá trị của tham số m bằng bao nhiêu để đồ thị hàm số y=x^4-2mx^2+1 có ba điểm cực trị A(0,1), B,C thỏa mãn BC=4 Câu 2: Các giá trị của tham s

$$\eqalign{
& Cau\,\,1: \cr
& y = {x^4} – 2m{x^2} + 1\,\,\left( {D = R} \right) \cr
& y’ = 4{x^3} – 4mx = 0 \Leftrightarrow 4x\left( {{x^2} – m} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 0 \Rightarrow y = 1 \Rightarrow A\left( {0;1} \right) \hfill \cr
{x^2} = m\,\,\left( * \right) \hfill \cr} \right. \cr
& De\,\,ham\,\,so\,\,co\,\,3\,\,cuc\,\,tri\,\,thi\,\,\left( * \right)\,\,co\,\,2\,\,nghiem\,\,pb \cr
& \Leftrightarrow m > 0 \cr
& \left( * \right) \Leftrightarrow \left[ \matrix{
x = \sqrt m \Rightarrow y = – {m^2} + 1 \hfill \cr
x = – \sqrt m \Rightarrow y = – {m^2} + 1 \hfill \cr} \right. \cr
& \Rightarrow B\left( {\sqrt m ; – {m^2} + 1} \right);\,\,C\left( { – \sqrt m ; – {m^2} + 1} \right) \cr
& BC = 4 \Leftrightarrow B{C^2} = 16 \cr
& \Leftrightarrow {\left( { – 2\sqrt m } \right)^2} = 16 \Leftrightarrow 4m = 16 \Leftrightarrow m = 4\,\,\left( {tm} \right) \cr
& Vay\,\,m = 4 \cr
& Cau\,\,2: \cr
& y = {x^4} – 2\left( {m + 1} \right){x^2} + {m^2}\,\,\left( {D = R} \right) \cr
& y’ = 4{x^3} – 4\left( {m + 1} \right)x = 0 \cr
& \Leftrightarrow 4x\left( {{x^2} – \left( {m + 1} \right)} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 0 \Rightarrow y = {m^2} \Rightarrow A\left( {0;{m^2}} \right) \hfill \cr
{x^2} = m + 1\,\,\left( * \right) \hfill \cr} \right. \cr
& De\,\,ham\,\,so\,\,co\,\,3\,\,cuc\,\,tri\,\,thi\,\,\left( * \right)\,\,co\,\,2\,\,nghiem\,\,pb \cr
& \Leftrightarrow m + 1 > 0 \Leftrightarrow m > – 1 \cr
& \left( * \right) \Leftrightarrow \left[ \matrix{
x = \sqrt {m + 1} \Rightarrow y = – 2m – 1 \hfill \cr
x = – \sqrt {m + 1} \Rightarrow y = – 2m – 1 \hfill \cr} \right. \cr
& \Rightarrow B\left( {\sqrt {m + 1} ; – 2m – 1} \right);\,\,C\left( { – \sqrt {m + 1} ; – 2m – 1} \right) \cr
& De\,\,thay\,\,\Delta ABC\,\,can\,\,tai\,\,A \cr
& \Rightarrow \Delta ABC\,\,vuong\,\,tai\,\,A \cr
& \Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 0 \cr
& \overrightarrow {AB} = \left( {\sqrt {m + 1} ; – 2m – 1 – {m^2}} \right);\,\,\overrightarrow {AC} = \left( { – \sqrt {m + 1} ; – 2m – 1 – {m^2}} \right) \cr
& \overrightarrow {AB} .\overrightarrow {AC} = 0 \cr
& \Leftrightarrow – \left( {m + 1} \right) + {\left( { – 2m – 1 – {m^2}} \right)^2} = 0 \cr
& \Leftrightarrow – \left( {m + 1} \right) + {\left( {m + 1} \right)^4} = 0 \cr
& \Leftrightarrow \left( {m + 1} \right)\left[ {{{\left( {m + 1} \right)}^3} – 1} \right] = 0 \cr
& \Leftrightarrow \left[ \matrix{
m + 1 = 0 \hfill \cr
m + 1 = 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
m = – 1\,\,\left( {loai} \right) \hfill \cr
m = 0\,\,\left( {tm} \right) \hfill \cr} \right. \cr
& Vay\,\,m = 0 \cr} $$