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\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
Nên \(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcz}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcz}{c^2}=\dfrac{abz-acy+bcx-abz+acy-bcz}{a^2+b^2+c^2}=\dfrac{0}{a^2+b^2+c^2}=0\)
Nên \(\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{c}{z}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(=\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)
\(=\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)
\(=\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}\)
\(=\dfrac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow abz-acy=bcx-abz=acy-bcx\)
\(\Rightarrow a\left(bz-cy\right)=b\left(cx-az\right)=c\left(ay-bx\right)\)
\(\Rightarrow bz-cy=cx-az=ay-bx\)
\(\Rightarrow\left\{{}\begin{matrix}bx=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{c}=\dfrac{y}{b}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{y}{b}=\dfrac{x}{a}\end{matrix}\right.\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
Vậy \(x:y:z=a:b:c\)
Lời giải:
Áp dụng TCDTSBN:
$\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}$
$=\frac{bza-cya}{a^2}=\frac{cxb-azb}{b^2}=\frac{ayc-bxc}{c^2}$
$=\frac{bza-cya+cxb-azb+ayc-bxc}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$
$\Rightarrow bz-cy=cx-az=ay-bx$
$\Rightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}$
Hay $a:b:c=x:y:z$ (đpcm)
Ta có :
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}\)
\(=\frac{ayz-bxz}{cz}=\frac{0}{ax+by+cz}=0\)
\(\Leftrightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\) \(\left(1\right)\)
\(cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\) \(\left(2\right)\)
\(ay=bx\Rightarrow\frac{y}{b}=\frac{x}{a}\) \(\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) hay \(x:y:z=a:b:c\)
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abx-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
\(=\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)
=>bz-cy=cx-az=ay-bx=0
- bz-cy=0 => bz=cy => \(\frac{b}{y}=\frac{c}{z}\)
- cx-az=0 => cx=az => \(\frac{c}{z}=\frac{a}{x}\)
=>\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\Rightarrow a:b:c=x:y:z\)(đpcm)
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=\dfrac{abz-acy}{a^2}=\dfrac{bcx-baz}{b^2}=\dfrac{cay-cbz}{c^2}=\dfrac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}\)
\(\Rightarrow\dfrac{bz-cy}{a}=0\)
\(\Rightarrow bz-cy=0\)
\(\Rightarrow bz=cy\Rightarrow\dfrac{b}{y}=\dfrac{c}{z}\)
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
Ta có \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Rightarrow\frac{\left(bz-cy\right).x}{ax}=\frac{\left(cx-az\right)y}{by}=\frac{\left(ay-bx\right).z}{cz}\)
\(\Rightarrow\frac{bxz-cxy}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}\)
Suy ra:
bz - cy = 0 (1)
cx - az = 0 (2)
ay - bx = 0 (3)
Từ (1) ta có: \(bz=cy\Rightarrow\frac{y}{b}=\frac{z}{c}\left(I\right)\)
Từ (2) ta có: \(cx=az=\frac{z}{c}=\frac{x}{a}\left(II\right)\)
Từ (3) ta có: \(ay=bx=\frac{x}{a}=\frac{y}{b}\left(III\right)\)
Từ (I), (II), (III) => x: y: z = a: b: c
Ta có: \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}=\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}=\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-abx}{c^2}\)
\(=\dfrac{abz-acy+bcx-abz+acy-abx}{a^2+b^2+c^2}\)
\(=\dfrac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow abz-acy=bcx-abz=acy-abx\)
\(\Rightarrow a\left(bz-cy\right)=b\left(cx-az\right)=c\left(ay-bx\right)\)
\(\Rightarrow bz-cy=cx-az=ay-bx\)
\(\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\dfrac{z}{c}=\dfrac{y}{b};\dfrac{x}{a}=\dfrac{z}{c};\dfrac{y}{b}=\dfrac{x}{a}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\Rightarrow x:y:z=a:b:c\)
Vậy x:y:z = a:b:c
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