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

=>x=ak+2bk+ck; y=2ak+bk-ck; z=4ak-4bk+ck

=> \(\frac{a}{x+2y+c}\)=\(\frac{a}{ak+2bk+ck+4bk+2bk-2ck+4ak-4bk+ck}\)=\(\frac{a}{9ak}\)=\(\frac{1}{9k}\)

Tương tự => \(\frac{a}{x+2y+c}\)=\(\frac{b}{2x+y-z}\)=\(\frac{c}{4x-4y+z}\)=\(\frac{1}{9k}\)

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!

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

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

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

\(\frac{4x}{4\left(a+2b+c\right)}=\frac{4y}{4\left(2a+b-c\right)}=\frac{z}{4a-4b+c}=\frac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}=\frac{4x-4y+z}{9c}\left(3\right)\)

Từ (1),(2),(3) => \(\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)

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

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

Cho xa+2b+c =y2a+b−c =z4a−4b+c 

Chứng minh : ax+2y+z =b2x+y−z =c4x−4y+z 

Toán lớp 7

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

xa+2b+c =2y2(2a+b−c) =z4a−4b+c =x+2y+za+2b+c+4a+2b−2c+4a−4b+c =x+2y+z9a (1)

2x2(a+2b+c) =y2a+b−c =z4a−4b+c =2x+y−z2a+4b+c+2a+b−c−4a+4b−c =2x+y−z9b (2)

4x4(a+2b+c) =4y4(2a+b−c) =z4a−4b+c =4x−4y+z4a+8b+4c−8a−4b+4c+4a−4b+c =4x−4y+z9c (3)

Từ (1),(2),(3) => x+2y+z9a =2x+y−z9b =4x−4y+z9c 

=> x+2y+za =2x+y−zb =4x−4y+zc 

=> ax+2y+z =b2x+y−z =c4x−4y+z 

Đơn giản thôi!!

Từ giả thiết, suy ra

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

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

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

Từ (1) , (2) và (3) suy ra:

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

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

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

Thằng này tự đăng tự làm cho đúng làm gì ???? ảo

Làm ơn bớt trẻ con  và suy nghĩ người lớn giùm cái, giỏi thì solo vs anh đây nè!