Violympic toán 7
Các câu hỏi tương tự
Cho \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
Chứng minh : \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
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\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y+z}\)
\(Cho\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\). CMR\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Cho : \(\dfrac{x}{a-2b+c}=\dfrac{y}{2a-b-c}=\dfrac{z}{4a+4b+c}\)
CM: \(\dfrac{a}{x+2y+z}=\dfrac{b}{z-y-2x}=\dfrac{c}{4z-4y+z}\)
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 dãy tỉ số bằng nhau:
\(\frac{x}{a+2b+c}=\frac{y}{2x+y-z}=\frac{z}{4a-4b+c}\)
Chứng minh rằng:\(\frac{a}{x+2y+z}=\frac{b}{2x+y-x}=\frac{c}{4x-4y+z}\)
Giúp mình với!
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 \(\dfrac{x}{2a+b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
CMR:\(\dfrac{a}{2x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)