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Ta có: \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x}{2a+4b+2c}=\frac{2y}{4a+2b-2c}\)
\(=\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{\left(a+2b+c\right)+\left(4a+2b-2c\right)+\left(4a-4b+c\right)}=\frac{x+2y+z}{9a}\left(1\right)\)
\(\frac{2x}{2a+4b+2c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x+y-z}{\left(2a+4b+2c\right)+\left(2a+b-c\right)-\left(4a-4b+c\right)}=\frac{2x+y-z}{9b}\left(2\right)\)
\(\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}=\frac{z}{4a-4b+c}=\frac{4x-4y+z}{\left(4a+8b+4c\right)-\left(8a+4b-4c\right)+\left(4a-4b+c\right)}=\frac{4x-4y+z}{9c}\left(2\right)\)
Từ (1); (2); (3) \(\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)
\(\Rightarrow\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}\)
\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right)\)
Ta có: \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}.\)
\(\Rightarrow\frac{2x}{2a+4b+2c}=\frac{2y}{4a+2b-2c}.\)
\(\Rightarrow\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
Từ \(\left(1\right),\left(2\right)và\left(3\right)\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}.\)
\(\Rightarrow\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}.\)
\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right).\)
Chúc bạn học tốt!
Ta co :
x/(a+2b+c)=y/(2a+b-c)=z/(4a-4b+c)
=> (a+2b+c)/x =(2a+b-c)/y =(4a-4b+c)/z
=>(a+2b+c)/x =2(2a+b-c)/2y =(4a-4b+c)/z
Ap dụng tính chất dãy tỉ số bang nhau ta có :
(a+2b+c)/x =(a+2b+c+4a+2b-2c+4a-4b+c)/(x+2y+z)
=>(a+2b+c)/x=9a/(x+2y+z)
Mặt khác (a+2b+c)/x =(2a+4b+2c+2a+b-c-4a+4b-c)/(2x+y-z)
=>(a+2b+c)/x =9b/(2x+y-z)
Làm tương tự cũng được (a+2b+c)/x =9c/(4x-4y+z) ==> ĐPCM
Ta có:
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2y}{4a+2b-2c}=\frac{x+2y+z}{9a}=\frac{1}{9}.\frac{x+2y+z}{a}\)(1)
\(=\frac{2x}{2a+4b+c}=\frac{2x+y-z}{9b}=\frac{1}{9}.\frac{2x+y-z}{b}\) (2)
\(=\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}=\frac{4x-4y+z}{9c}=\frac{1}{9}.\frac{4x-4y+z}{c}\) (3)
Từ (1), (2) và (3) => \(\frac{1}{9}.\frac{x+2y+z}{a}=\frac{1}{9}.\frac{2x+y-z}{b}=\frac{1}{9}.\frac{4x-4y+z}{c}\)
=> \(\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}\)
=> \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
Ta có: \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x}{2a+4b+2c}=\dfrac{2y}{4a+4b-2c}=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+y+z}{\left(a+2b+c\right)+\left(2a+b-c\right)+\left(4a-4b+c\right)}=\dfrac{x+2y+z}{9b}\left(1\right)\)
\(\dfrac{2x}{2a+2b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-z}{\left(2a+2b+2c\right)+\left(2a+b-c\right)-\left(4a-4b+c\right)}=\dfrac{2x+y-z}{9a}\left(2\right)\)
\(\dfrac{4x}{4a+4b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{\left(4a+8b+4c\right)-\left(8a+4b-4c\right)+\left(4a-4b+c\right)}=\dfrac{4x-4y+z}{9c}\left(3\right)\)
Từ (1), (2), (3) \(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y+z}{9b}=\dfrac{4x-4y+z}{9b}\)
\(\Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\left(đpcm\right)\)
Chúc bạn học tốt!
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