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