tìm GTLN của A biết a,b>0 và a thuộc N* : A= (2√a+ √b)/(√ab+2√a- √b-2)-(2√a-√b)/(√ab+2√a+√b+2)
giúp mik nha
0 bình luận về “tìm GTLN của A biết a,b>0 và a thuộc N* : A= (2√a+ √b)/(√ab+2√a- √b-2)-(2√a-√b)/(√ab+2√a+√b+2)
giúp mik nha”
Giải thích các bước giải:
Ta có:
\(\begin{array}{l} A = \left( {\dfrac{{2\sqrt a + \sqrt b }}{{\sqrt {ab} + 2\sqrt a – \sqrt b – 2}}} \right) – \left( {\dfrac{{2\sqrt a – \sqrt b }}{{\sqrt {ab} + 2\sqrt a + \sqrt b + 2}}} \right)\\ = \dfrac{{2\sqrt a + \sqrt b }}{{\sqrt a \left( {\sqrt b + 2} \right) – \left( {\sqrt b + 2} \right)}} – \dfrac{{2\sqrt a – \sqrt b }}{{\sqrt a \left( {\sqrt b + 2} \right) + \left( {\sqrt b + 2} \right)}}\\ = \dfrac{{2\sqrt a + \sqrt b }}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)}} – \dfrac{{2\sqrt a – \sqrt b }}{{\left( {\sqrt a + 1} \right)\left( {\sqrt b + 2} \right)}}\\ = \dfrac{{\left( {2\sqrt a + \sqrt b } \right)\left( {\sqrt a + 1} \right) – \left( {2\sqrt a – \sqrt b } \right)\left( {\sqrt a – 1} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{2a + 2\sqrt a + \sqrt {ab} + \sqrt b – \left( {2a – 2\sqrt a – \sqrt {ab} + \sqrt b } \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{4\sqrt a + 2\sqrt {ab} }}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{2\sqrt a \left( {\sqrt b + 2} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{2\sqrt a }}{{\left( {\sqrt a – 1} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{2\sqrt a }}{{a – 1}} \end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \left( {\dfrac{{2\sqrt a + \sqrt b }}{{\sqrt {ab} + 2\sqrt a – \sqrt b – 2}}} \right) – \left( {\dfrac{{2\sqrt a – \sqrt b }}{{\sqrt {ab} + 2\sqrt a + \sqrt b + 2}}} \right)\\
= \dfrac{{2\sqrt a + \sqrt b }}{{\sqrt a \left( {\sqrt b + 2} \right) – \left( {\sqrt b + 2} \right)}} – \dfrac{{2\sqrt a – \sqrt b }}{{\sqrt a \left( {\sqrt b + 2} \right) + \left( {\sqrt b + 2} \right)}}\\
= \dfrac{{2\sqrt a + \sqrt b }}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)}} – \dfrac{{2\sqrt a – \sqrt b }}{{\left( {\sqrt a + 1} \right)\left( {\sqrt b + 2} \right)}}\\
= \dfrac{{\left( {2\sqrt a + \sqrt b } \right)\left( {\sqrt a + 1} \right) – \left( {2\sqrt a – \sqrt b } \right)\left( {\sqrt a – 1} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2a + 2\sqrt a + \sqrt {ab} + \sqrt b – \left( {2a – 2\sqrt a – \sqrt {ab} + \sqrt b } \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{4\sqrt a + 2\sqrt {ab} }}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2\sqrt a \left( {\sqrt b + 2} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt b + 2} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2\sqrt a }}{{\left( {\sqrt a – 1} \right)\left( {\sqrt a + 1} \right)}}\\
= \dfrac{{2\sqrt a }}{{a – 1}}
\end{array}\)