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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: \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x}{2a+4b+2c}=\frac{2y}{4a+2b-2c}\)
\(=\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{x}{a+2b+c}=\frac{2y}{4a+2b-2c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{\left(a+2b+c\right)+\left(4a+2b-2c\right)+\left(4a-4b+c\right)}=\frac{x+2y+z}{9a}\left(1\right)\)
\(\frac{2x}{2a+4b+2c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{2x+y-z}{\left(2a+4b+2c\right)+\left(2a+b-c\right)-\left(4a-4b+c\right)}=\frac{2x+y-z}{9b}\left(2\right)\)
\(\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}=\frac{z}{4a-4b+c}=\frac{4x-4y+z}{\left(4a+8b+4c\right)-\left(8a+4b-4c\right)+\left(4a-4b+c\right)}=\frac{4x-4y+z}{9c}\left(2\right)\)
Từ (1); (2); (3) \(\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)
\(\Rightarrow\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}\)
\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right)\)
Ta có: \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}.\)
\(\Rightarrow\frac{2x}{2a+4b+2c}=\frac{2y}{4a+2b-2c}.\)
\(\Rightarrow\frac{4x}{4a+8b+4c}=\frac{4y}{8a+4b-4c}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
Từ \(\left(1\right),\left(2\right)và\left(3\right)\Rightarrow\frac{x+2y+z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}.\)
\(\Rightarrow\frac{x+2y+z}{a}=\frac{2x+y-z}{b}=\frac{4x-4y+z}{c}.\)
\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right).\)
Chúc bạn học tốt!
Ta co :
x/(a+2b+c)=y/(2a+b-c)=z/(4a-4b+c)
=> (a+2b+c)/x =(2a+b-c)/y =(4a-4b+c)/z
=>(a+2b+c)/x =2(2a+b-c)/2y =(4a-4b+c)/z
Ap dụng tính chất dãy tỉ số bang nhau ta có :
(a+2b+c)/x =(a+2b+c+4a+2b-2c+4a-4b+c)/(x+2y+z)
=>(a+2b+c)/x=9a/(x+2y+z)
Mặt khác (a+2b+c)/x =(2a+4b+2c+2a+b-c-4a+4b-c)/(2x+y-z)
=>(a+2b+c)/x =9b/(2x+y-z)
Làm tương tự cũng được (a+2b+c)/x =9c/(4x-4y+z) ==> ĐPCM
\(CMR:\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
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 \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\left(đpcm\right)\)