Phương pháp:
\(\displaystyle {a \over b} = {c \over d} \Rightarrow {a \over c} = {b \over d}\)
\(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a - c}}{{b - d}}\,\left( {b,d,b - d \ne 0} \right)\)
Lời giải:
a) \(\displaystyle {a \over b} = {c \over d} \Rightarrow {a \over c} = {b \over d}\)
\(\displaystyle \Rightarrow {{ab} \over {cd}}= {a \over c}.{b \over d} = {a \over c}.{a \over c} = {b \over d}.{b \over d} \)\(\,\displaystyle = {{{a^2} - {b^2}} \over {{c^2} - {d^2}}}\)
b) \(\displaystyle {a \over b} = {c \over d} \Rightarrow {a \over c} = {b \over d} = {{a - b} \over {c - d}} \)
\(\displaystyle \Rightarrow {{ab} \over {cd}} = {a \over c}.{b \over d} = {{a - b} \over {c - d}}.{{a - b} \over {c - d}} \)\(\,\displaystyle = {{{{\left( {a - b} \right)}^2}} \over {{{\left( {c - d} \right)}^2}}}\)
Bài 8.5
Tìm \(x, y\) biết: \(\displaystyle {2 \over x} = {3 \over y}\) và \(xy = 96\).
Phương pháp:
\(\dfrac{a}{b} = \dfrac{c}{d}\)\(\, \Rightarrow {\left( {\dfrac{a}{b}} \right)^2} = \dfrac{a}{b}.\dfrac{c}{d}\,\left( {b,d \ne 0} \right)\)
Lời giải:
Từ \(\displaystyle {2 \over x} = {3 \over y}\) ta có \(\displaystyle {4 \over {{x^2}}} = {2 \over x}.{3 \over y} = {6 \over {xy}} = {6 \over {96}} = {1 \over {16}} \)
\( \Rightarrow \dfrac{4}{{{x^2}}} = \dfrac{1}{{16}} \Rightarrow {x^2} = 4.16 = 64\)
\(\Rightarrow x = \pm 8\)
- Nếu \(x = 8\) thì \(y = 96 : 8 = 12\).
- Nếu \(x = -8\) thì \(y = 96 : (-8) = -12\).
Bài 8.6
Biết rằng \(\displaystyle {{bz - cy} \over a} = {{cx - az} \over b} = {{ay - bx} \over c}.\)
Hãy chứng minh \(x : y : z = a : b : c\).
Phương pháp:
\(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = \dfrac{{a + c + e}}{{b + d + f}}\)\(\,\,\left( {b,d,f,b + d + f \ne 0} \right)\)
Lời giải:
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:
\(\displaystyle bz = cy \Rightarrow {z \over c} = {y \over b}\) (1)
\(\displaystyle cx = az \Rightarrow {x \over a} = {z \over c}\) (2)
\(\displaystyle ay = bx \Rightarrow {y \over b} = {x \over a}\) (3)
Từ (1), (2), (3) suy ra \(\displaystyle {x \over a} = {y \over b} = {z \over c}\) hay \(x : y : z = a : b : c.\)
