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ta có: \(\frac{x^2-yz}{a}=\frac{y^2-xz}{b}=\frac{z^2-xy}{c}\)
\(\Rightarrow\frac{a}{x^2-yz}=\frac{b}{y^2-xz}=\frac{c}{z^2-xy}\Rightarrow\frac{a^2}{\left(x^2-yz\right)^2}=\frac{b^2}{\left(y^2-xz\right)^2}=\frac{c^2}{\left(z^2-xy\right)^2}\) (1)
=> \(\frac{a}{\left(x^2-yz\right)}.\frac{a}{\left(x^2-yz\right)}=\frac{b}{y^2-xz}.\frac{c}{z^2-xy}=\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-xz\right).\left(z^2-xy\right)}\)
a^2/(x^2-yz)^2 = (a^2-bc)/[(x^2-yz)^2 - (y^2-xz)(z^2-xy)] = (a^2-bc)/[x (x^3 + y^3 + z^3 - 3xyz)] =>
(a^2-bc)/x = [a^2/(x^2 - yz)^2] * (x^3 + y^3 + z^3 - 3xyz) (2)
Thực hiện tương tự ta cũng có
(b^2-ac)/y = [b^2/(y^2 - xz)^2] * (x^3 + y^3 + z^3 - 3xyz) (3)
(c^2-ab)/z = [c^2/(z^2 - xy)^2] * (x^3 + y^3 + z^3 - 3xyz) (4)
Từ (1),(2),(3),(4) => (a^2-bc)/x = (b^2-ac)/y = (c^2-ab)/z.
\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{xz}{cx+az}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)
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!
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
mà\(\frac{xy}{ay+bx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\)\(\frac{atbt}{abt+bat}=\frac{a^2t^2+b^2t^2+c^2t^2}{a^2+b^2+c^2}\)
\(\Rightarrow\frac{t}{2}=t^2\Rightarrow t=\frac{1}{2}\)
\(\Rightarrow t=\frac{1}{2}\Rightarrow\hept{\begin{cases}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{cases};\left(a,b,c\ne0\right)}\)
với x=y=z khác 0 và a,b,c khác nhau là 1 số bất kỳ khác 0 thì (1) thỏa mãn và (2) không thỏa mãn
=> Không thể CM
ta có: \(\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\)
\(\Rightarrow\frac{a}{x^2-yz}=\frac{b}{y^2-zx}=\frac{c}{z^2-xy}\) (*)
\(\Rightarrow\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-zx\right).\left(z^2-xy\right)}=\frac{a^2-bc}{\left(x^2-yz\right)^2-\left(y^2-zx\right).\left(z^2-xy\right)}\)
\(=\frac{a^2-bc}{x^4-3x^2yz+xy^3+xz^3}=\frac{a^2-bc}{x.\left(x^3-3xyz+y^3+z^3\right)}\)
\(\Rightarrow\frac{a^2-bc}{x}=\frac{a^2}{\left(x^2-yz\right)^2}.\left(x^3-3xyz+y^3+z^3\right)\)
Làm tương tự như trên. ta có:
\(\frac{b^2-ca}{y}=\frac{b^2}{\left(y^2-zx\right)^2}.\left(x^3-3xyz+y^3+z^3\right)\)
\(\frac{c^2-ab}{z}=\frac{c^2}{\left(z^2-xy\right)^2}.\left(x^3-3xyz+y^3+z^3\right)\)
Từ (*) \(\Rightarrow\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\left(đpcm\right)\)
Thử tiếp này \(\frac{a}{x^2-yz}=\frac{b}{y^2-xz}=\frac{c}{z^2-xy}\)
=> \(\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-xz\right)\left(z^2-xy\right)}=\frac{a^2-bc}{\left(x^2-yz\right)^2-\left(y^2-xz\right)\left(z^2-xy\right)}\)
Có \(\frac{x^2-yz}{a}=\frac{y^2-xz}{b}=\frac{z^2-xy}{c}\)
=> \(\frac{a}{x^2-yz}=\frac{b}{y^2-xz}=\frac{c}{z^2-xy}\)
=> \(\frac{a^2}{\left(x^2-yz\right)^2}=\frac{bc}{\left(y^2-xz\right).\left(z^2-xy\right)}=\frac{a^2-bc}{\left(x^2-yz\right)^2-\left(y^2-xz\right).\left(z^2-xy\right)}\)
\(=\frac{b^2}{\left(y^2-xz\right)^2}=\frac{ac}{\left(x^2-yz\right).\left(z^2-xy\right)}=\frac{b^2-ac}{\left(y^2-xz\right)^2-\left(x^2-yz\right).\left(z^2-xy\right)}\)
\(=\frac{c^2}{\left(z^2-xy\right)^2}=\frac{ab}{\left(x^2-yz\right).\left(y^2-xz\right)}=\frac{c^2-ab}{\left(z^2-xy\right)^2-\left(x^2-yz\right).\left(y^2-xz\right)}\)
Xét (x2 - yz)2 - (y2 - xz)(z2 - xy)
= ...................... (Tui xét phía dưới rùi kéo xuống phía dưới mà coi)
= x(x3 + y3 + z3 - 3xyz)
Tương tự, ta có (y2-xz)2 - (x2 - yz).(z2 - xy) = y.(x3 + y3 + z3 - 3xyz)
(z2 - xy)2 - (x2 - yz).(y2 - xz) = z.(x3 + y3 + z3 - 3xyz)
=> \(\frac{a^2-bc}{x\left(x^2+y^3+z^3-3xyz\right)}=\frac{b^2-ac}{y\left(x^3+y^3+z^3-3xyz\right)}=\frac{c^2-ab}{z\left(x^3+y^3+z^3-3xyz\right)}\)
=> \(\frac{a^2-bc}{x}=\frac{b^2-ac}{y}=\frac{c^2-ab}{z}\)(Đpcm)