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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 => \(\frac{a}{x+2y+c}\)=\(\frac{b}{2x+y-z}\)=\(\frac{c}{4x-4y+z}\)=\(\frac{1}{9k}\)
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)
\(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{x}{4a-4b+6}\) 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
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=\frac{x+2y+z}{9a}\)( 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 ) và ( 3 )
\(\frac{x+2y-z}{9a}=\frac{2x+y-z}{9b}=\frac{4x-4y+z}{9c}\)hay \(\frac{9a}{x+2y-z}=\frac{9b}{2x+y-z}=\frac{9c}{4x-4y+z}\)
\(\Rightarrow\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
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
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