Bạn xem lời giải  Tại đây  nhé !

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!

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

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

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

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

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

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

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

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

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

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

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

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

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

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

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

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

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

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

\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)(vì tất cả các tử và mẫu khác 0)

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

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

Áp dụng TC DTSBN ta có :

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

\(=\frac{x+2y+z}{9a}=\frac{1}{9}.\frac{x+2y+z}{a}\) (1)

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

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

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

\(=\frac{4x-4y+z}{9c}=\frac{1}{9}.\frac{4x-4y+z}{c}\)(3)

Từ (1);(2);(3) \(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y+z}=\frac{c}{4x-4y+z}\) (đpcm)

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

Sửa đề trong bài làm luôn nhé

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

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

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

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

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

Từ (1), (2), (3) ta có ĐPCM

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

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

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

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

Từi (1),(2),(3) 

còn j giải típ nha

@@@@@@@@@@@@

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