Ta có :

\(\dfrac{cy-bx}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}=\dfrac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}=0\)

\(\Rightarrow\dfrac{cy-bz}{x}=0\) \(\Rightarrow cy=bz\) \(\Rightarrow\) \(\dfrac{b}{y}=\dfrac{c}{z}\left(1\right)\)

\(\Rightarrow\dfrac{az-cx}{y}=0\) \(\Rightarrow az=cx\) \(\Rightarrow\dfrac{a}{x}=\dfrac{c}{z}\left(2\right)\)

Từ (1) và (2) suy ra : \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

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

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

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\dfrac{a\left(bz-cy\right)}{a^2}\text{=}\dfrac{b\left(cx-az\right)}{b^2}\text{=}\dfrac{c\left(ay-bx\right)}{c^2}\text{=}\dfrac{abz-acy+bcz-baz+cay-cbx}{a^2+b^2+c^2}\text{=}0\)

\(\Rightarrow\dfrac{bz-cy}{a}\text{=}0\Rightarrow bz\text{=}cy\)

\(\Rightarrow\dfrac{b}{c}\text{=}\dfrac{y}{z}\left(1\right)\)

\(\dfrac{cx-az}{b}\text{=}0\Rightarrow cx\text{=}az\)

\(\Rightarrow\dfrac{c}{a}\text{=}\dfrac{z}{x}\left(2\right)\)

Từ (1) và (2):

\(\Rightarrow dpcm\)

Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\) thì \(x=ak,y=bk,z=ck\)

\(\dfrac{bz-cy}{a}=\dfrac{bck-bck}{a}=0\) __( 1 )__

\(\dfrac{cx-az}{b}=\dfrac{ack-ack}{b}=0\) __( 2 )__

\(\dfrac{ay-bx}{c}=\dfrac{abk-abk}{c}=0\) __( 3 )__

Từ ( 1 ), ( 2 ), ( 3 ) suy ra \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

Lời giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

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

\(\Rightarrow \left\{\begin{matrix} bz-cy=0\\ cx-az=0\\ ay-bx=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} bz=cy\\ cx=az\\ ay=bx\end{matrix}\right.\)

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

Do đó ta có đpcm.

\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=\dfrac{abz-acy}{a^2}=\dfrac{bcx-baz}{b^2}=\dfrac{cay-cbz}{c^2}=\dfrac{abz-acy+bcx-baz+cay-cbz}{a^2+b^2+c^2}\)

\(\Rightarrow\dfrac{bz-cy}{a}=0\)

\(\Rightarrow bz-cy=0\)

\(\Rightarrow bz=cy\Rightarrow\dfrac{b}{y}=\dfrac{c}{z}\)

tương tự \(\dfrac{c}{z}=\dfrac{a}{x}\)

Vậy \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)

đặt x=ak, y=bk, z=ck

thay vào biểu thức là ra mà

\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=\dfrac{abz-acy}{a^2}=\dfrac{bcx-baz}{b^2}=\dfrac{cay-cbz}{c^2}=\dfrac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}\)

\(\Rightarrow\dfrac{bz-cy}{a}=0\)

\(\Rightarrow bz-cy=0\)

\(\Rightarrow bz=cy\Rightarrow\dfrac{b}{y}=\dfrac{c}{z}\)

tương tự \(\dfrac{c}{z}=\dfrac{a}{x}\)

Vậy \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)

Câu hỏi của Huyền Trang Tiến Tài - Toán lớp 7 | Học trực tuyến

Lời giải:
Sửa đề: $z$ đầu tiên ở mẫu đổi thành $a$.

Ta có:

$\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}$

$=\frac{abz-cya}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}$

$=\frac{abz-cya+bcx-abz+acy-bcx}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$

$\Rightarrow bz-cy=cx-az=ay-bx=0$

$\Rightarrow bz=cy; cx=az; ay=bx$

$\Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

Ta có đpcm.