Tính góc A của tam giác ABC thoả mãn hệ thức b(b^2-a^2 ) = c( a^2. - c^2 )

C xen thử đi

b^3-ba^2=a^2c-c^3

< = > (b+c)(b^2-bc+c^2)-a^2(b+c)=0

< = > (b+c)(b^2-bc+c^2-a^2)=0

< = > b^2-bc+c^2=a^2=b^2+c^2-2bc.cosA

< = > bc(2cosA-1)=0

< = > cosA=1/2

< = > góc A=60

\[\begin{array}{l} b\left( {{b^2} - {a^2}} \right) = c\left( {{c^2} - {a^2}} \right)\\ \Leftrightarrow {b^3} - b{a^2} = {c^3} - c{a^2}\\ \Leftrightarrow {b^3} - {c^3} - \left( {b{a^2} - c{a^2}} \right) = 0\\ \Leftrightarrow \left( {b - c} \right)\left( {{b^2} + bc + {c^2}} \right) - {a^2}\left( {b - c} \right) = 0\\ \Leftrightarrow \left( {b - c} \right)\left( {{b^2} + bc + {c^2} - {a^2}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} b - c = 0\\ {b^2} + bc + {c^2} = {a^2} \end{array} \right.\,\,\,\\ + )\,\,\,b - c = 0 \Leftrightarrow b = c \Rightarrow \Delta ABC\,\,can\,\,\,tai\,\,A.\\ + )\,\,{b^2} + bc + {c^2} = {a^2} \Leftrightarrow {b^2} + {c^2} - {a^2} = - bc\\ Ta\,\,co:\,\,\,\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{ - bc}}{{2bc}} = - \frac{1}{2}\\ \Rightarrow \cos A = - \frac{1}{2} \Rightarrow \angle A = {120^0}. \end{array}\]