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Giải:
Đặt \(A=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
Ta có:
\(A=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\dfrac{x+2y+z}{9a}\)
\(A=\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-x}{2a+4b+2c+2a+b-c-4a+4b-c}=\dfrac{2x+y-x}{9b}\)
\(A=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{4a+8b-8a-4b+4c+4a-4b+c}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow A=\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
\(\RightarrowĐPCM\)
Đặt \(k=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4b-4a-c}\)
Do đó: \(k=\dfrac{x}{a+2b+c}=\dfrac{2y}{4a+2b-2c}=\dfrac{z}{4b-4a-c}\)
\(k=\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4b-4a-c}\)
\(k=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4b-4a-c}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(k=\dfrac{x+2y-z}{a+2b+c+4a+2b-2c-4b+4a+c}=\dfrac{x+2y-z}{9a}\)
\(k=\dfrac{2x+y+z}{2a+4b+2c+2a+b-a+4b-4a-c}=\dfrac{2x+y+z}{9b}\)
\(k=\dfrac{4x-4y-z}{4a+8b+4c-8a-4b+4c-4b+4a+c}=\dfrac{4x-4y-z}{9c}\)
\(\Rightarrow\dfrac{x+2y-z}{9a}=\dfrac{2x+y+z}{9b}=\dfrac{4x-4y-z}{9c}\)
\(\Rightarrow\dfrac{x+2y-z}{a}=\dfrac{2x+y+z}{b}=\dfrac{4x-4y-z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y-z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y-z}\) => đpcm
Đặt A= \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
Ta có:
\(A=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\dfrac{x+2y+z}{9a}\)
\(A=\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b-c}=\dfrac{2x+y-z}{9b}\)\(A=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}=\dfrac{4x-4y+z}{9c}\)\(\Rightarrow A=\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Đặt \(\dfrac{a}{x}\)=\(\dfrac{b}{y}\)=\(\dfrac{c}{z}\)=m
\(\Rightarrow\)a=xm ; b=ym ; c=zm
Thay a=xm ; b=ym ; c=zm vào \(\dfrac{ak^2+bk+c}{xk^2+yk+z}\)ta có:
\(\dfrac{ak^2+bk+c}{xk^2+yk+z}\)=\(\dfrac{xmk^2+ymk+zm}{xk^2+yk+z}\)=\(\dfrac{m\left(xk^2+yk+z\right)}{xk^2+yk+z}\)=m
\(\Rightarrow\)đpcm
