Cho x^2-yz/a=y^2-zx/b=z^2-xy/c Chứng minh rằng a^2-bc/x=b^2-ca/y=c^2-ab/z
Những câu hỏi liên quan
cho \(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\). Chứng minh rằng \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
Cho a,b,c,x,y,z khác 0 thỏa mãn : \(\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}=k\).Chứng minh rằng: \(\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\)
Cho \(\frac{x^2-yz}{a}=\frac{y^2-zx}{b}=\frac{z^2-xy}{c}\) Chứng minh : \(\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\)
Cho a, b, x, y, z là các số khác 0 thỏa mãn: \(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\ne0\). CMR: \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
Cho các số thực a,b,c,x,y,z khác 0 thỏa mãn
\(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\)
Chứng minh rằng \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ac}{y}=\dfrac{c^2-ab}{z}\)
Cho \(\frac{x^2-yz}{a}=\frac{y^2-xz}{b}=\frac{z^2-xy}{c}\). Chứng minh rằng: \(\frac{a^2-bc}{x}=\frac{b^2-ca}{y}=\frac{c^2-ab}{z}\)
Cho a,b, c, x, y, z là các sô thực dương thỏa mãn điều kiện x+ y+z =1. Chứng minh
rằng:
\(ax+by+cz+2\sqrt{\left(xy+yz+zx\right)\left(ab+bc+ca\right)}\le a+b+c\)
Cho a; b; c; x; y; z và \(x^2-yz\ne0;y^2-zx\ne0;z^2-xy\ne0\) thỏa mãn \(\dfrac{x^2-yz}{a}=\dfrac{y^2-xz}{b}=\dfrac{z^2-xy}{c}\). CMR \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
ta có:(a+b)(b+c)(c+a)(a+b+c)(ab+bc+ca)-abcgeleft(a+b+cright)left(ab+bc+caright)-frac{1}{9}left(a+b+cright)left(ab+bc+caright)frac{8}{9}left(a+b+cright)left(ab+bc+caright)frac{x}{x+yz}+frac{y}{y+zx}+frac{z}{z+xy}frac{x}{left(x+yright)left(x+zright)}+frac{y}{left(y+xright)left(y+zright)}+frac{z}{left(z+xright)left(z+yright)}frac{2left(xy+yz+zxright)}{left(x+yright)left(y+zright)left(z+xright)}lefrac{9}{4left(xy+yz+zxright)}frac{9}{4}
Đọc tiếp
ta có:
(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{1}{9}\left(a+b+c\right)\left(ab+bc+ca\right)=\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}=\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+x\right)\left(y+z\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\frac{9}{4\left(xy+yz+zx\right)}=\frac{9}{4}\)