Toán Lớp 9: Giúp mik bài 1 vs ạ????
1 A=√15 – √12/√5-2 – 6+2√6/√3 +√2
B=(3√x-1/x-1 – 1/√x-1): 1/x+√x
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: Giúp mik bài 1 vs ạ????
1 A=√15 – √12/√5-2 – 6+2√6/√3 +√2
B=(3√x-1/x-1 – 1/√x-1): 1/x+√x
Comments ( 1 )
A = \dfrac{{\sqrt {15} – \sqrt {12} }}{{\sqrt 5 – 2}} – \dfrac{{6 + 2\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }}\\
= \dfrac{{\sqrt 3 .\sqrt 5 – \sqrt 4 .\sqrt 3 }}{{\sqrt 5 – 2}} – \dfrac{{\sqrt 2 .\sqrt 3 .\sqrt 6 + \sqrt 2 .\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 5 – 2} \right)}}{{\sqrt 5 – 2}} – \dfrac{{\sqrt 2 .\sqrt 6 \left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }}\\
= \sqrt 3 – \sqrt 2 .\sqrt 6 \\
= \sqrt 3 – 2\sqrt 3 \\
= – \sqrt 3 \\
B = \left( {\dfrac{{3\sqrt x – 1}}{{x – 1}} – \dfrac{1}{{\sqrt x – 1}}} \right):\dfrac{1}{{x + \sqrt x }}\\
= \dfrac{{3\sqrt x – 1 – \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}.\sqrt x \left( {\sqrt x + 1} \right)\\
= \dfrac{{2\sqrt x – 2}}{{\sqrt x – 1}}.\sqrt x \\
= \dfrac{{2\left( {\sqrt x – 1} \right)}}{{\sqrt x – 1}}.\sqrt x \\
= 2\sqrt x
\end{array}$