Có:

\(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)

\(\Rightarrow ax^2+by^2+cz^2\)

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

\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)

Mà \(a+b+c=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)

Đồng thời có: \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

Từ đây ta có:)

\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)

\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)

\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)

Giả sử điều cần c/m là đúng . Khi đó , ta có :

\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)

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

\(\Leftrightarrow x^2a^2+y^2a^2+z^2a^2+x^2b^2+y^2b^2+z^2b^2+x^2c^2+y^2c^2+z^2c^2\)

\(=x^2a^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2=2axby+2bycz+2axcz\)

\(\Leftrightarrow y^2a^2+z^2a^2+x^2b^2+z^2b^2+x^2c^2+y^2c^2-2axby-2bycz-2axcz=0\) \(\Leftrightarrow\left(y^2a^2-2axby+b^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)+\left(x^2c^2-2axcz+a^2z^2\right)=0\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2=0\left(1\right)\)

Do \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\\\left(cx-az\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(ay-bx\right)^2+\left(bz-cy\right)^2+\left(cx-az\right)^2\ge0\left(2\right)\)

Từ ( 1 ) ; ( 2 ) \(\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\bz-cy=0\\cx-az=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\bz=cy\\cx=az\end{matrix}\right.\)

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

Điều này đúng với GT đề bài cho

\(\Rightarrow\) Điều cần c/m là đúng

\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\left(đpcm\right)\)

Đặt x/a=y/b=z/c=k

=>x=ak; y=bk; z=ck

\(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{a^2k^2+b^2k^2+c^2k^2}{a^4k^2+b^4k^2+c^4k^2}=\dfrac{1}{a^2+b^2+c^2}\)

Đặt \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{1}{k}\Rightarrow x=ak;y=bk;y=ck\)

\(\Rightarrow\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{a^2k^2+b^2k^2+c^2k^2}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{k^2\left(a^2+b^2+c^2\right)}{k^2\left(a^2+b^2+c^2\right)^2}=\frac{1}{a^2+b^2+c^2}\)

Mạo phép sửa đề!CMR: \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{3}{a^2+b^2+c^2}\)

Ta có: \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) 

\(\Rightarrow\frac{x^2}{ax}=\frac{y^2}{by}=\frac{z^2}{cz}=\frac{x^2+y^2+z^2}{ax+by+cz}\)  (t/c dãy tỉ số bằng nhau)

\(\Rightarrow\frac{x^2}{\left(ax\right)^2}=\frac{y^2}{\left(by\right)^2}=\frac{z^2}{\left(cz\right)^2}=\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}\) (1)

Lại có: \(\frac{x^2}{\left(ax\right)^2}=\frac{y^2}{\left(by\right)^2}=\frac{z^2}{\left(cz\right)^2}=\) \(\frac{x^2}{a^2x^2}=\frac{y^2}{b^2y^2}=\frac{z^2}{c^2z^2}=\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{3}{a^2+b^2+c^2}\)

Chết,sai cmnr....=((

Cộng vế với vế:

\(\Rightarrow x+y+z=2ax+2by+2cz\)

\(\Rightarrow x+y+z-2x=2ax+2by+2cx-2\left(by+cz\right)=2ax\)

\(\Rightarrow2ax=y+z-x\)

\(\Rightarrow a=\dfrac{y+z-x}{2x}\Rightarrow1+a=\dfrac{x+y+z}{2x}\)

Tương tự ta có: \(1+b=\dfrac{x+y+z}{2y}\) ; \(1+c=\dfrac{x+y+z}{2z}\)

\(\Rightarrow\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}=\dfrac{2x+2y+2z}{x+y+z}=2\)

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

Nguyễn Huy Tú Lightning Farron Akai Haruma

ta có x+y+z=0 =>x^2=(y+z)^2
y^2=(x+z)^2
z^2=(x+y)^2
do đó ax^2+by^2+cz^2
=a(y+z)^2+b(x+z)^2+c(x+y)^2
=a(y^2+2yz+z^2)+b(x^2+2xz+z^2)
+c(x^2+2xy+y^2)
=x^2(b+c)+y^2(a+c)+z^2(a+b)
+2(ayz+bxz+cxy) (1)
thay b+c=-a ,a+c=-b , a+b=-c do a+b+c=0
và ayz+bxz+cxy=0 do a/x+b/y+c/z=0 vào (1) ta được
ax^2+by^2+cz^2 = -(ax^2+by^2+cz^2)
=> ax^2+by^2+cz^2=0

Ta có:

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{a+b}{x+y}=0\Leftrightarrow ay\left(x+y\right)+bx\left(x+y\right)+xy\left(a+b\right)=0\)

\(\Leftrightarrow axy+ay^2+bx^2+bxy+axy+bxy=0\)

\(\Leftrightarrow ay^2+2axy+2bxy+bx^2=0\)

Vậy:

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

\(=ax^2+by^2-\left(ax^2+2axy+ay^2+bx^2+2bxy+by^2\right)\)

\(=-\left(ay^2+2axy+2bxy+by^2\right)=-0=0\)