Cho \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) khác 0. Tính giá trị của \(\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{ax^2+by^2+cz^2}\)
Cho \(\dfrac{bz+cy}{x\left(-ax+by+cz\right)}=\dfrac{cx+az}{y\left(ax-by+cz\right)}=\dfrac{ay+bx}{z\left(ax+by-cz\right)}\)
CMR : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
b) \(\dfrac{x}{a\left(b^2+c^2-a^2\right)}=\dfrac{y}{b\left(a^2+c^2-b^2\right)}=\dfrac{z}{c\left(a^2+b^2-c^2\right)}\)
Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng
Nguyễn Huy Tú Lightning Farron Akai Haruma
Cho: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)và x, y, z khác 0
CMR: \(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
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: \(\frac{x^2}{\left(ax\right)^2}=\frac{y^2}{\left(by\right)^2}=\frac{z^2}{\left(cz\right)^2}=\) \(\frac{x^2}{a^2x^2}=\frac{y^2}{b^2y^2}=\frac{z^2}{c^2z^2}=\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{3}{a^2+b^2+c^2}\)
cho \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) rut gon\(\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^22}\)
Cho x,y,z,a,b,c khác 0 và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\).Chứng minh rằng
a) \(\dfrac{a^2}{x}=\dfrac{b^2}{y}=\dfrac{c^2}{z}=\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
b) \(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)}=\dfrac{1}{a^2+b^2+c^2}\)
C/m rằng nếu \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\) với x,y,z khác 0 thì \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
Trừ cả 2 vế cho \(a^2x^2+b^2y^2+c^2z^2\), ta có:
\(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2axby+2bycz+2axcz\)
\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2axcz=0\)
\(\left(a^2y^2+b^2x^2-2axby\right)+\left(a^2z^2+c^2z^2-2axcz\right)+\left(b^2z^2+c^2y^2-2bycz\right)=0\)
\(\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay-bx=0\\az-cx=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
=> đpcm
Cho ax+by+cz=0; a+b+c=\(\dfrac{1}{100}\); ax2+by2+cz2 khác 0. Tính\(S=\dfrac{\text{ax^2+by^2+cz^2}}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}\)
Cho x, y, z khác 0 và a, b, c dương thoả mãn ax+by+cz=0 và a+b+c=2017. Tính giá trị của biểu thức: \(P=\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}\)
\(Cho\) \(ax+by+cz=0;a+b+c=\dfrac{1}{2018}\) . CMR: \(\dfrac{ax^{2\:}+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}=2018\)
Cho ax+by+cz=0; a+b+c =\(\dfrac{2019}{2018}\)
Tính : \(P=\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}\)
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