Biết rằng \(\displaystyle {{bz - cy} \over a} = {{cx - az} \over b} = {{ay - bx} \over c}.\)
Hãy chứng minh \(x : y : z = a : b : c\).
Hướng dẫn giải
\(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = \dfrac{{a + c + e}}{{b + d + f}}\)\(\,\,\left( {b,d,f,b + d + f \ne 0} \right)\)
Lời giải chi tiết
Ta có:
\(\displaystyle {{bz - cy} \over a} = {{cx - az} \over b} = {{ay - bx} \over c} \)
\(\displaystyle= {{bxz - cxy} \over {ax}} = {{cxy - ayz} \over {by}} = {{ayz - bxz} \over {cz}}\)
\(=\dfrac{{bxz - cxy + cxy - ayz + ayz - bxz}}{{ax + by + cz}}\)
\(\displaystyle= {0 \over {ax + by + cz}} = 0\)
Suy ra:
\(+)\,\dfrac{{bz - cy}}{a} = 0 \Rightarrow bz - cy = 0\)\(\displaystyle \Rightarrow bz = cy \Rightarrow {z \over c} = {y \over b}\) (1)
\(+)\,\dfrac{{cx - az}}{b} = 0 \Rightarrow cx - az = 0\)\(\displaystyle \Rightarrow cx = az \Rightarrow {x \over a} = {z \over c}\) (2)
\(+)\,\dfrac{{ay - bx}}{c} = 0 \Rightarrow ay - bx = 0\)\(\displaystyle \Rightarrow ay = bx \Rightarrow {y \over b} = {x \over a}\) (3)
Từ (1), (2), (3) suy ra \(\displaystyle {x \over a} = {y \over b} = {z \over c}\) hay \(x : y : z = a : b : c.\)
-- Mod Toán 7