câu 1)
câu a )[tex]P=\frac{\left [(a+1)^2+3(a-1)^2 \right ](a+1)^2}{\left [(a-1)^2+3(a+1)^2 \right ](a-1)^2}.\frac{(a-1)(a^2+a+1)}{(a+1)(a^2-a+1)}-\frac{2a}{a-1}=\frac{4(a^2-a+1)(a+1)^2.(a-1)(a^2+a+1)}{4(a^2+a+1)(a-1)^2(a+1)(a^2-a+1)}-\frac{2a}{a-1}=\frac{a+1}{a-1}-\frac{2a}{a-1}=-1[/tex]
câu b ) [tex]\sqrt{x^2(1+\sqrt[3]{\frac{y^2}{x^2}})}+\sqrt{y^2(1+\sqrt[3]{\frac{x^2}{y^2}})}=a<=>\sqrt{(\sqrt[3]{x^2}+\sqrt[3]{y^2})}\left ( \sqrt[3]{x^2}+\sqrt[3]{y^2} \right )=a<=>\left ( \sqrt[3]{x^2}+\sqrt[3]{y^2} \right )^3=a^2<=>\left ( \sqrt[3]{x^2}+\sqrt[3]{y^2} \right )=\sqrt[3]{a^2}[/tex]
câu 3).
câu a) vì 1 và a là 2 no của pt [tex]P(x)[/tex]
[tex](x-1)(x-a)=x^2+ax+b<=>x^2-(a+1)x+a=x^2+ax+b[/tex]
đồng nhất hệ số ta được : [tex]\left\{\begin{matrix} -(a+1)=a=>a=\frac{-1}{2} & \\ a=b& \end{matrix}\right.[/tex]
=> [tex]a=b=\frac{-1}{2}[/tex]
câu b)
[tex](x_{2}-x_{1})^2=(x_{3}-x_{4})^2<=>(x_{1}+x_{2})^2-4x_{2}x_{1}=(x_{3}+x_{4})^2-4x_{3}x_{4}[/tex]
áp dụng viet : [tex]\left\{\begin{matrix} x_{1}+x_{2}=\frac{1}{2} & & & \\ x_{1}x_{2}=\frac{-1}{2} & & & \\ x_{3}+x_{4}=-c & & & \\ x_{3}x_{4}=d & & & \end{matrix}\right.[/tex]
[tex]x_{3}^2+x_{4}^2+a(x_{3}+x_{4})+2b=x_{1}^2+x_{2}^2+c(x_{1}+x_{2})+2d<=>(x_{3}-x_{4})^2-ac+2b+2d=(x_{1}-x_{2})^2-ac+2d+2b<=>(x_{1}-x_{2})^2=(x_{3}-x_{4})^2[/tex]