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 \(\left(1\right),\left(2\right)và\left(3\right)\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}.\)

\(\Rightarrow\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}.\)

\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right).\)

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vào đây nhé bạn

\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)

=>\(\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}\)

Áp dụng t/c dãy tỉ số=nhau:

\(\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}\)\(=\frac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\frac{x+2y+z}{9a}\left(1\right)\)

\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)

=>\(\frac{2x}{2a+4b+2c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)

Áp dụng t/c dãy tỉ số=nhau:

\(\frac{2x}{2a+4b+2c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\)\(\frac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b-c}=\frac{2x+y-z}{9b}\left(2\right)\)

\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)

=>\(\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}=\frac{z}{4a-4b+c}\)

Áp dụng t/c dãy tỉ số=nhau:

\(\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}=\frac{z}{4a-4b+c}=\)\(\frac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}=\frac{4x-4y+z}{9c}\left(3\right)\)

Từ (1);(2);(3) ta có dãy tỉ số=nhau:

\(\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)

=>\(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\) (đpcm)

(*) bài này thiếu dữ kiện:" giả thiết các tỉ số đều có nghĩa" nhé,phải có dữ liệu đó mới suy ra đpcm được

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) \(\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)

\(\Rightarrow\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}\)

\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right)\)

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{x}{4a-4b+6}\) thì \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y+z}\)

Giải:

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+2y+z}{9a}\left(1\right)\)

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-z}{9b}\left(2\right)\)

\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{9c}\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)hay

\(\dfrac{a}{x+2y+z}=\dfrac{b}{2z+y-z}=\dfrac{c}{4x-4y+z}\) cùng = 9