Toán Lớp 9: (6 – 2√5)x = √5 – 1
b²x² – b² = x² (b>1)
ax + 3 = √3 x + a² ( a≠√3)
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Toán Lớp 9: (6 – 2√5)x = √5 – 1
b²x² – b² = x² (b>1)
ax + 3 = √3 x + a² ( a≠√3)
Giúp mình với. Mình cảm ơn
Comments ( 1 )
a)\left( {6 – 2\sqrt 5 } \right)x = \sqrt 5 – 1\\
\Leftrightarrow {\left( {\sqrt 5 – 1} \right)^2}.x = \sqrt 5 – 1\\
\Leftrightarrow x = \dfrac{{\sqrt 5 – 1}}{{{{\left( {\sqrt 5 – 1} \right)}^2}}} = \dfrac{1}{{\sqrt 5 – 1}}\\
\Leftrightarrow x = \dfrac{{\sqrt 5 + 1}}{{\left( {\sqrt 5 – 1} \right)\left( {\sqrt 5 + 1} \right)}} = \dfrac{{\sqrt 5 + 1}}{4}\\
Vậy\,x = \dfrac{{\sqrt 5 + 1}}{4}\\
b)b.{x^2} – {b^2} = {x^2}\left( {b > 1} \right)\\
\Leftrightarrow b.{x^2} – {x^2} = {b^2}\\
\Leftrightarrow \left( {b – 1} \right).{x^2} = {b^2}\\
\Leftrightarrow {x^2} = \dfrac{{{b^2}}}{{b – 1}}\\
\Leftrightarrow x = \dfrac{b}{{\sqrt {b – 1} }} = \dfrac{{b\sqrt {b – 1} }}{{b – 1}}\\
Vậy\,x = \dfrac{{b\sqrt {b – 1} }}{{b – 1}}\\
c)a.x + 3 = \sqrt 3 x + {a^2}\left( {a\# \sqrt 3 } \right)\\
\Leftrightarrow \left( {a – \sqrt 3 } \right).x = {a^2} – 3\\
\Leftrightarrow \left( {a – \sqrt 3 } \right).x = \left( {a – \sqrt 3 } \right)\left( {a + \sqrt 3 } \right)\\
\Leftrightarrow x = a + \sqrt 3 \\
Vậy\,x = a + \sqrt 3
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