Question

a) Cho \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)

CM: \(\frac{a}{x+2y+z}=\frac{b}{2x+y+z}=\frac{c}{4x-4b+c}\)

b) Cho \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

CM: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

Answer 1

a) Đề sai nhé !

b) Ta có :  \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(\Rightarrow\frac{abz-cya}{a^2}=\frac{bcx-abz}{b^2}=\frac{cay-bcx}{c^2}=\frac{abz-cya+bcx-abz+cay-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow abz-cya=0\Leftrightarrow abz=cya\Leftrightarrow bz=cy\Leftrightarrow\frac{y}{b}=\frac{z}{c}\)(1)

\(\Rightarrow bcx-abz=0\Leftrightarrow bcx=abz\Leftrightarrow cx=az\Leftrightarrow\frac{x}{a}=\frac{z}{c}\)(2)

Từ (1) và (2) ta có \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

Answer 2

ok. Thank you. Ổn rồi

Answer 3

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Answer 4

đơ đó bn