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\(\frac{x}{a-2b+c}=\frac{y}{2a-b-c}=\frac{z}{4a+4b+c}\)
\(=\frac{2y}{4a-2b-2c}=\frac{2x}{2a-4b+2c}=\frac{4x}{4a-8b+4c}=\frac{4y}{8a-4b-4c}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{a-2b+c}=\frac{2y}{4a-2b-2c}=\frac{z}{4a+4b+c}=\frac{x+2y+z}{9a}\left(1\right)\)
\(\frac{z}{4a+4b+c}=\frac{y}{2a-b-c}=\frac{2x}{2a-4b+2c}=\frac{z-y-2x}{9b}\left(2\right)\)
\(\frac{4x}{4a-8b+4c}=\frac{4y}{8a-4b-4c}=\frac{z}{4a+4b+c}=\frac{4x-4y+z}{9c}\left(3\right)\)
Từ (1),(2),(3) \(\Rightarrow\frac{x+2y+z}{9a}=\frac{z-y-2x}{9b}=\frac{4x-4y+z}{9c}\) \(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{z-y-2x}=\frac{x}{4x-4y+z}\)(ĐPCM)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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Áp dụng t/c của DTSBN , ta có :
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{x+2y+z}{a+2b+c+2\left(2a+b-c\right)+4a-4b+c}\\ =\dfrac{x+2y+z}{a+2b+c+4a+2b-2a-2c+4a-4b+c}\\ =\dfrac{x+2y+z}{\left(a+4a+4a\right)+\left(2b+2b-4b\right)+\left(c-2c+c\right)}\\ =\dfrac{x+2y+z}{9a}\left(1\right)\)
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{2x+y-z}{2\left(a+2b+c\right)+2a+b-c-4a+4b+c}\\ =\dfrac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b+c}\\ =\dfrac{2x+y-z}{\left(2a+2a-4a\right)+\left(4b+b+4b\right)+\left(2c-c+c\right)}\\ =\dfrac{2x+y-z}{9b}\left(2\right)\)
+, \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\\ =\dfrac{4x-4y+z}{4\left(a+2b+c\right)-4\left(2a+b-c\right)++4a-4b+c}\\ =\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}\\ =\dfrac{4x-4y+z}{\left(4a-8a+4a\right)+\left(8b-4b-4b\right)+\left(4c+4c+c\right)}\\ =\dfrac{4x-4y+z}{9c}\left(3\right)\)
Từ (1);(2) và (3)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2a+y-z}{9b}=\dfrac{4x-4y+z}{9c}\\ \Rightarrow\dfrac{x+2y+z}{9a}\cdot9=\dfrac{2a+y-z}{9b}\cdot9=\dfrac{4x-4y+z}{9c}\cdot9\\ \Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2a+y-z}{b}=\dfrac{4x-4y+z}{c}\\ \Rightarrow\dfrac{a}{a+2y+z}=\dfrac{b}{2a+y-z}=\dfrac{c}{4x-4y+z}\left(đpcm\right)\)
Đặt \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=k\left(a+2b+c\right)\\y=k\left(2a+b-c\right)\\z=k\left(4a-4b+c\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{a}{k\left(a+2b+c\right)+2k\left(2a+b-c\right)+k\left(4a-4b+c\right)}=\dfrac{a}{k.9a}=\dfrac{1}{9k}\)
Tượng tự:
\(\dfrac{b}{2x+y-z}=\dfrac{b}{9bk}=\dfrac{1}{9k}\) ; \(\dfrac{c}{4x-4y+z}=\dfrac{c}{9k.c}=\dfrac{1}{9k}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn lưu ý, gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề của bạn hơn nhé.
Sửa đề c/m : \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
Ta có \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
=> \(\frac{a+2b+c}{x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}\)
Từ (1) => \(\frac{a+2b+c}{x}=\frac{4a+2b-2c}{2y}=\frac{4a-4b+c}{z}=\frac{a+2b+c+4a+2b-2c+4a-4b+c}{x+2y+z}\)
\(=\frac{9a}{x+2y+z}\)(2)
Từ (1) => \(\frac{2a+4b+2c}{2x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}=\frac{2a+4b+2c+2a+b-c-4a+4b-c}{2x+y-z}\)
\(=\frac{9b}{2x+y-z}\)(3)
Từ (1) => \(\frac{4a+8b+4c}{4x}=\frac{8a+4b-4c}{4y}=\frac{4a-4b+c}{z}\)
\(=\frac{4a+8a+4c-8a-4b+4c+4a-4b+c}{4x-4y+z}=\frac{9c}{4x-4y+z}\)(4)
Từ (2)(3)(4) => \(\frac{9a}{x+2y+z}=\frac{9b}{2x+y-z}=\frac{9c}{4x-4y+z}\)
=> \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)(đpcm)