1.Cho x+y+z=0 ,rút gọn:
\(A=\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
2.Tính \(A=\dfrac{x-y}{x+y}\)biết x2-2y2=xy (y khácx;x+y khác 0)
Cho x,y,z khác 0 và \(\dfrac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
CMR:\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
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
