a: \(\dfrac{xy}{x^2+y^2}=\dfrac{5}{8}\)

=>\(\dfrac{xy}{5}=\dfrac{x^2+y^2}{8}=k\)

=>\(xy=5k;x^2+y^2=8k\)

\(A=\dfrac{8k-2\cdot5k}{8k+2\cdot5k}=\dfrac{-2}{18}=\dfrac{-1}{9}\)

b: Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\)

=>x=a*k; y=b*k; z=c*k

\(B=\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{\left(a\cdot ak+b\cdot bk+c\cdot ck\right)^2}\)

\(=\dfrac{k^2\cdot\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)

Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\) \(\left(k\ne0\right)\) \(\Rightarrow\left\{{}\begin{matrix}x=a.k\\y=b.k\\z=c.k\end{matrix}\right.\)

Ta có :

\(A=\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)

\(A=\dfrac{\left[\left(a.k\right)^2+\left(b.k\right)^2+\left(c.k\right)^2\right]\cdot\left(a^2+b^2+c^2\right)}{\left(a.a.k+b.b.k+c.c.k\right)^2}\)

\(A=\dfrac{\left(a^2k^2+b^2k^2+c^2k^2\right)\cdot\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)

\(A=1\)

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

1) Đặt \(B=x^2+y^2+z^2\)

\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)

Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)

\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)

Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)

\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)

2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)

\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)

\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)

Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)

Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)

3) Từ giả thiết, ta suy ra

\(\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)

Bạn chịu khó tự biến đổi, ta được

\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)

\(\left\{{}\begin{matrix}ay-bx=0\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}\\az-cx=0\Leftrightarrow az=cx\Leftrightarrow\dfrac{a}{x}=\dfrac{c}{z}\\bz-cy=0\Leftrightarrow bz=cy\Leftrightarrow\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\)

Vậy ta suy ra đpcm

P/S: Đây là BĐT Bunyakovsky với 3 số