K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

27 tháng 2 2019

Ta có:

\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+az}\left(x;y;z\ne0\right)\)

=> \(\frac{xyz}{azy+bxz=}=\frac{xyz}{xbz+xcy}=\frac{yzx}{ycx+azy}\)

=>\(zay+bxz=xbz+xyc=ycx+azy\)

\(\Rightarrow\hept{\begin{cases}za=cx\\bx=ay\end{cases}}\)

Đặt \(\frac{x}{a}=\frac{z}{c}=\frac{y}{b}=t\left(t\ne0\right)\)

=> x = at ; z = ct  ; y = bt

\(\frac{xy}{ay+bx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\)\(\frac{atbt}{abt+bat}=\frac{a^2t^2+b^2t^2+c^2t^2}{a^2+b^2+c^2}\)

\(\Rightarrow\frac{t}{2}=t^2\Rightarrow t=\frac{1}{2}\)

\(\Rightarrow t=\frac{1}{2}\Rightarrow\hept{\begin{cases}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{cases};\left(a,b,c\ne0\right)}\)

25 tháng 2 2019

Câu hỏi của Hacker Chuyên Nghiệp:tham khảo

10 tháng 1 2020

\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{xz}{cx+az}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)

Ta có: \(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{xz}{cx+az}.\)

\(\Rightarrow\frac{xyz}{ayz+bxz}=\frac{xyz}{bxz+cxy}=\frac{xyz}{cxy+ayz}.\)

\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)

\(\Rightarrow\left\{{}\begin{matrix}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ayz=cxy\\bxz=cxy\\bxz=ayz\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}az=cx\\bz=cy\\bx=ay\end{matrix}\right.\left(2\right)\)

Thay (2) vào (1) ta được:

\(\frac{xy}{ay+ay}=\frac{yz}{bz+bz}=\frac{xz}{cx+cx}\)

\(\Rightarrow\frac{xy}{2ay}=\frac{yz}{2bz}=\frac{xz}{2cx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right).\)

\(\Rightarrow\frac{x^2}{4a^2}=\frac{y^2}{4b^2}=\frac{z^2}{4c^2}=\frac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\frac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)

\(\Rightarrow\frac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}=\frac{1.\left(x^2+y^2+z^2\right)}{4.\left(a^2+b^2+c^2\right)}\)

\(\Rightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{1}{4}\left(4\right).\)

Từ (3) và (4)

\(\Rightarrow\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{1}{4}.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2a}=\frac{1}{4}\\\frac{y}{2b}=\frac{1}{4}\\\frac{z}{2c}=\frac{1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{4}.2a\\y=\frac{1}{4}.2b\\z=\frac{1}{4}.2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{matrix}\right.\)

Vậy \(x=\frac{a}{2};y=\frac{b}{2};z=\frac{c}{2}\left(x,y,z\ne0\right);\left(a,b,c\ne0\right).\)

Chúc bạn học tốt!

17 tháng 12 2018

\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+az}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Leftrightarrow\frac{x}{a}+\frac{y}{b}=\frac{y}{b}+\frac{z}{c}=\frac{z}{c}+\frac{x}{a}\)

\(\hept{\begin{cases}\frac{x}{a}+\frac{y}{b}=\frac{y}{b}+\frac{z}{c}\Rightarrow\frac{x}{a}=\frac{z}{c}\\\frac{z}{c}+\frac{x}{a}=\frac{y}{b}+\frac{z}{c}\Rightarrow\frac{x}{a}=\frac{y}{b}\\\frac{x}{a}+\frac{y}{b}=\frac{z}{c}+\frac{x}{a}\Rightarrow\frac{y}{b}=\frac{z}{c}\end{cases}}\Rightarrow\frac{x}{a}=\frac{z}{c}=\frac{y}{b}.\text{đăt}k=\frac{x}{a}=\frac{z}{c}=\frac{y}{b}\Rightarrow x=ak,z=ck,y=bk\)

ta có: \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{k^2.\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)}=k^2\Rightarrow k^2=2k\Rightarrow k^2-2k=0\Rightarrow k.\left(k-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}k=0\\k=2\end{cases}\text{mà a,b,c và x,y,z khác 0. }\Rightarrow k=2\Rightarrow x=2a,y=2b,z=2c}\)

p/s: bài nì khó chơi vc =.=" sai sót bỏ qua ^^'