Đáp án + Giải thích các bước giải:
Bài 2:
$(a^2+b^2+c^2)(x^2+y^2+z^2)=(ax+by+cz)^2\\\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2axby+2axcz+2abycz\\\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2-a^2x^2-b^2y^2-c^2z^2-2axby-2axcz-2abycz=0\\\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\\\Leftrightarrow (ay-bx)^2+(bz-cy)^2+(az-cx)^2=0\\\Leftrightarrow \begin{cases}(ay-bx)^2=0\\(bz-cy)^2=0\\(az-cx)^2=0\end{cases}\\\Leftrightarrow \begin{cases}ay-bx=0\\bz-cy=0\\az-cx=0\end{cases}\\\Leftrightarrow \begin{cases}ay=bx\\bz=cy\\az=cx\end{cases}\\\Rightarrow \begin{cases}\dfrac ax=\dfrac by\\\dfrac by=\dfrac cz\\\dfrac cz=\dfrac ax\end{cases}\\\Rightarrow \dfrac ax=\dfrac by=\dfrac cz$
Bài 3:
Ta có: $(a+b)^2=2(a^2+b^2)\\\Leftrightarrow a^2+2ab+b^2=2a^2+2b^2\\\Leftrightarrow -a^2+2ab-b^2=0\\\Leftrightarrow -(a^2-2ab+b^2)=0\\\Leftrightarrow -(a-b)^2=0\\\Leftrightarrow a-b=0\\\Leftrightarrow a=b$