Question

[Vũ Thị Cúc]

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

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

Bạn nào làm được giải giúp mình nha.....giải chi tiết vào nha.....mình đang cần gấp....mai kiểm tra....

Answer 1

[Unknown_Hacker]

Có: Đề \(\Leftrightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)\(=\frac{\left(abz-abz\right)+\left(bcx-bcx\right)+\left(acy-acy\right)}{a^2+b^2+c^2}\)

\(=\frac{0}{a^2+b^2+c^2}=0\)\(\left(ĐKXĐ:a,b,c\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Leftrightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}}\RightarrowĐpcm\)

Answer 2

[Nguyễn Thị Lan Hương]

\(\frac{bz-cy}{a}\)=\(\frac{cx-az}{b}\)=\(\frac{ay-bx}{c}\)=>\(\frac{a\left(bz-cy\right)}{a^2}\)=\(\frac{b\left(cx-az\right)}{b^2}\)=\(\frac{c\left(ay-bx\right)}{c^2}\)

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

=>\(\frac{bz-cy}{a}\)=\(\frac{cx-az}{b}\)=\(\frac{ay-bx}{c}\)= 0

=> bz - cy = cx - az = ay - bx = 0

+) bz - cy = 0 => bz = cy => y/b = z/c

+) cx - az = 0 => cx = az => x/a = z/c

=> x/a = y/b = z/c