Cho \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
C/m rằng: \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\) với các mẫu số ≠ 0
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Cho\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
C/m rằng:\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\) với các mẫu số ≠ 0
Cho \(\dfrac{x}{\text{a}+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4b-4a-c}\)
Chứng minh rằng: \(\dfrac{a}{x+2y-z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y-z}\)
Đặt \(k=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4b-4a-c}\)
Do đó: \(k=\dfrac{x}{a+2b+c}=\dfrac{2y}{4a+2b-2c}=\dfrac{z}{4b-4a-c}\)
\(k=\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4b-4a-c}\)
\(k=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4b-4a-c}\)
Theo t/c dãy tỉ số bằng nhau, ta có:
\(k=\dfrac{x+2y-z}{a+2b+c+4a+2b-2c-4b+4a+c}=\dfrac{x+2y-z}{9a}\)
\(k=\dfrac{2x+y+z}{2a+4b+2c+2a+b-a+4b-4a-c}=\dfrac{2x+y+z}{9b}\)
\(k=\dfrac{4x-4y-z}{4a+8b+4c-8a-4b+4c-4b+4a+c}=\dfrac{4x-4y-z}{9c}\)
\(\Rightarrow\dfrac{x+2y-z}{9a}=\dfrac{2x+y+z}{9b}=\dfrac{4x-4y-z}{9c}\)
\(\Rightarrow\dfrac{x+2y-z}{a}=\dfrac{2x+y+z}{b}=\dfrac{4x-4y-z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y-z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y-z}\) => đpcm
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Chứng minh rằng :\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\) thì \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\).
cho \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\).CMR\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y+z}=\dfrac{c}{4x-4y+z}\)
\(Cho\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\). CMR\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Chứng minh rằng:\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}\dfrac{z}{4a-4b+c}\)
Thì:\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+x}\)
Đặt A= \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
Ta có:
\(A=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\dfrac{x+2y+z}{9a}\)
\(A=\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-z}{2a+4b+2c+2a+b-c-4a+4b-c}=\dfrac{2x+y-z}{9b}\)\(A=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{4a+8b+4c-8a-4b+4c+4a-4b+c}=\dfrac{4x-4y+z}{9c}\)\(\Rightarrow A=\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
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chứng minh rằng:
Nếu:\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)thì \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Giải:
Đặt \(A=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\)
Ta có:
\(A=\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{x+2y+z}{a+2b+c+4a+2b-2c+4a-4b+c}=\dfrac{x+2y+z}{9a}\)
\(A=\dfrac{2x}{2a+4b+2c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}=\dfrac{2x+y-x}{2a+4b+2c+2a+b-c-4a+4b-c}=\dfrac{2x+y-x}{9b}\)
\(A=\dfrac{4x}{4a+8b+4c}=\dfrac{4y}{8a+4b-4c}=\dfrac{z}{4a-4b+c}=\dfrac{4x-4y+z}{4a+8b-8a-4b+4c+4a-4b+c}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow A=\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{9a}=\dfrac{2x+y-z}{9b}=\dfrac{4x-4y+z}{9c}\)
\(\Rightarrow\dfrac{x+2y+z}{a}=\dfrac{2x+y-z}{b}=\dfrac{4x-4y+z}{c}\)
\(\Rightarrow\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
\(\RightarrowĐPCM\)
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17 tháng 11 2021 lúc 20:20
1.
a) CMR: Nếu a+b+c=0 thì \(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}=0\)
b) Nếu \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\) thì:
\(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+2y-z}=\dfrac{c}{4x-4y+z}\)
2. Cho \(\dfrac{x}{x^2+x+1}=a\) .Tính \(M=\dfrac{x^2}{x^4-x^2+1}\)
Cho \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\) Thì \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{\text{4x-4y+z}}\) HELP MEEEEEEEEEEEEE
https://hoc24.vn/hoi-dap/question/217666.html
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