Cho \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
Chứng minh : \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
CMR
Nếu \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
thì \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y=z}\)
(với điều kiện các tỉ số đều có nghĩa )
cho a,b,c,x,y,z thỏa mãn
\(\frac{x}{a+2b+c}=\frac{b}{2a+b-c}=\frac{z}{4a-4b+c}\)
CM:\(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
--\(Cho\frac{a}{b}=\frac{3}{4}.TínhA=\frac{a^2+3b^2}{a^2-3b^2}\)
--Cho\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
CMR \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
Please HELP meeeeeee🙏 🙏 🙏 🙏
1,cho\(\frac{2y+2z-x}{a}=\frac{2z+2x-y}{b}=\frac{2x+2y-z}{c}\)
CMR:\(\frac{x}{2b+2c-a}=\frac{y}{2c+2a-b}=\frac{z}{2a+2b-c}\)
Tìm các số x, y, z biết rằng:
a) \(\frac{y+z+1}{x}=\frac{x+y+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\);
b) \(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}\);
c) \(\frac{2x+1}{5}=\frac{3y-2}{7}=\frac{2x+3y-1}{6x}\)
Cho \(\frac{2y+2z-x}{a}=\frac{2z+2x-y}{b}=\frac{2x+2y-z}{c}\) với a, b,c khác 0; 2a+2b khác c; 2b+2c khác a; 2c+2a khác b.
CM: \(\frac{x}{2b+2c-a}=\frac{y}{2c+2a-b}=\frac{z}{2a+2b-c}\)
Chứng minh rằng:\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}\dfrac{z}{4a-4b+c}\)
Thì:\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+x}\)
cho \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\).CMR\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y+z}\)
