Bài 16 trang 30 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Chứng minh :

a) \(\left( {\dfrac{{2\sqrt 3  – \sqrt 6 }}{{\sqrt 8  – 2}} – \dfrac{{\sqrt {216} }}{3}} \right).\dfrac{1}{{\sqrt 6 }} =  – \dfrac{3}{2}\);       

b) \(\dfrac{{a\sqrt b  + b\sqrt a }}{{\sqrt {ab} }}:\dfrac{1}{{\sqrt a  – \sqrt b }} = a – b\)\(\;\left( {a,b > 0,a \ne b} \right)\).

Lời giải chi tiết

\(\begin{array}{l}a)\;\left( {\dfrac{{2\sqrt 3  – \sqrt 6 }}{{\sqrt 8  – 2}} – \dfrac{{\sqrt {216} }}{3}} \right).\dfrac{1}{{\sqrt 6 }} =  – \dfrac{3}{2}\\VT = \left( {\dfrac{{2\sqrt 3  – \sqrt 6 }}{{\sqrt 8  – 2}} – \dfrac{{\sqrt {216} }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\\;\;\;\;\; = \left( {\dfrac{{{{\left( {\sqrt 2 } \right)}^2}.\sqrt 3  – \sqrt 6 }}{{\sqrt {{2^2}.2}  – 2}} – \dfrac{{\sqrt {{6^2}.6} }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\\;\;\;\;\; = \left( {\dfrac{{\sqrt 6 \left( {\sqrt 2  – 1} \right)}}{{2\left( {\sqrt 2  – 1} \right)}} – \dfrac{{6\sqrt 6 }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\\;\;\;\;\; = \left( {\dfrac{{\sqrt 6 }}{2} – 2\sqrt 6 } \right).\dfrac{1}{{\sqrt 6 }} =  – \dfrac{{3\sqrt 6 }}{2}.\dfrac{1}{{\sqrt 6 }} =  – \dfrac{3}{2}.\\ \Rightarrow VT = VP.\end{array}\)                                  \(\begin{array}{l}b)\;\dfrac{{a\sqrt b  + b\sqrt a }}{{\sqrt {ab} }}:\dfrac{1}{{\sqrt a  – \sqrt b }} = a – b\left( {a,b > 0,a \ne b} \right)\\VT = \dfrac{{\sqrt {ab} \left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt {ab} }}:\dfrac{1}{{\sqrt a  – \sqrt b }}\\\;\;\;\;\; = \left( {\sqrt a  + \sqrt b } \right).\left( {\sqrt a  – \sqrt b } \right)\\\;\;\;\;\; = a – b.\\ \Rightarrow VT = VP.\end{array}\)

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