Répondre :
Bonsoir,
[tex](1-k)(-1-k)+(4k^2-9)\times\dfrac{1}{9}k^2-8=0\\\\(-1-k+k+k^2)+(\dfrac{4k^4}{9}-k^2)-8=0\\\\\dfrac{4k^4}{9}-9=0\\\\\dfrac{4k^4-81}{9}=0\\\\4k^4-81=0\\\\(2k^2-9)(2k^2+9)=0\\\\k\sqrt{2}-3)(k\sqrt{2}+3)(2k^2+9)=0[/tex]
[tex]k\sqrt{2}-3=0\ \ ou\ \ k\sqrt{2}+3=0\ \ ou\ \ 2k^2+9=0\\\\k\sqrt{2}-3=0\Longrightarrow k=\dfrac{3}{\sqrt{2}}\\\\k\sqrt{2}+3=0\Longrightarrow k=\dfrac{-3}{\sqrt{2}}\\\\2k^2+9=0\Longrightarrow\ impossible\ \ car\ \ 2k^2+9>0[/tex]
D'où :
[tex]S=\{\dfrac{3}{\sqrt{2}};-\dfrac{3}{\sqrt{2}}\}[/tex]
[tex](1-k)(-1-k)+(4k^2-9)\times\dfrac{1}{9}k^2-8=0\\\\(-1-k+k+k^2)+(\dfrac{4k^4}{9}-k^2)-8=0\\\\\dfrac{4k^4}{9}-9=0\\\\\dfrac{4k^4-81}{9}=0\\\\4k^4-81=0\\\\(2k^2-9)(2k^2+9)=0\\\\k\sqrt{2}-3)(k\sqrt{2}+3)(2k^2+9)=0[/tex]
[tex]k\sqrt{2}-3=0\ \ ou\ \ k\sqrt{2}+3=0\ \ ou\ \ 2k^2+9=0\\\\k\sqrt{2}-3=0\Longrightarrow k=\dfrac{3}{\sqrt{2}}\\\\k\sqrt{2}+3=0\Longrightarrow k=\dfrac{-3}{\sqrt{2}}\\\\2k^2+9=0\Longrightarrow\ impossible\ \ car\ \ 2k^2+9>0[/tex]
D'où :
[tex]S=\{\dfrac{3}{\sqrt{2}};-\dfrac{3}{\sqrt{2}}\}[/tex]