Toán Lớp 9: Thực hiện phép tính
a. 5+2 căn5/căn5+căn2
b.Căn(2-căn3/2+căn3)
c.(2/căn3-1 + 3/căn3-2 + 15/3-căn3) x 1/căn3+5
d.(căn14-căn7/1-căn2 + căn15-căn5/1-căn3) : 1/căn7-căn5
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Toán Lớp 9: Thực hiện phép tính
a. 5+2 căn5/căn5+căn2
b.Căn(2-căn3/2+căn3)
c.(2/căn3-1 + 3/căn3-2 + 15/3-căn3) x 1/căn3+5
d.(căn14-căn7/1-căn2 + căn15-căn5/1-căn3) : 1/căn7-căn5
Comments ( 2 )
a)\dfrac{{5 + 2\sqrt 5 }}{{\sqrt 5 + \sqrt 2 }} = \dfrac{{\left( {5 + 2\sqrt 5 } \right)\left( {\sqrt 5 – \sqrt 2 } \right)}}{{5 – 2}}\\
= \dfrac{{5\sqrt 5 – 5\sqrt 2 + 10 – 2\sqrt {10} }}{3}\\
b)\sqrt {\dfrac{{2 – \sqrt 3 }}{{2 + \sqrt 3 }}} = \sqrt {\dfrac{{{{\left( {2 – \sqrt 3 } \right)}^2}}}{{{2^2} – 3}}} = \sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} = 2 – \sqrt 3 \\
c)\left( {\dfrac{2}{{\sqrt 3 – 1}} + \dfrac{3}{{\sqrt 3 – 2}} + \dfrac{{15}}{{3 – \sqrt 3 }}} \right).\dfrac{1}{{\sqrt 3 + 5}}\\
= \left( {\dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 – 1}} + \dfrac{{3\left( {\sqrt 3 + 2} \right)}}{{3 – {2^2}}} + \dfrac{{15\left( {3 + \sqrt 3 } \right)}}{{{3^2} – 3}}} \right).\dfrac{{5 – \sqrt 3 }}{{25 – 3}}\\
= \left( {\sqrt 3 + 1 – 3\sqrt 3 – 6 + \dfrac{{15 + 5\sqrt 3 }}{2}} \right).\dfrac{{5 – \sqrt 3 }}{{22}}\\
= \dfrac{{\sqrt 3 + 5}}{2}.\dfrac{{5 – \sqrt 3 }}{{22}}\\
= \dfrac{{{5^2} – 3}}{{2.22}}\\
= \dfrac{1}{2}\\
d)\left( {\dfrac{{\sqrt {14} – \sqrt 7 }}{{1 – \sqrt 2 }} + \dfrac{{\sqrt {15} – \sqrt 5 }}{{1 – \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 – \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 7 \left( {\sqrt 2 – 1} \right)}}{{1 – \sqrt 2 }} + \dfrac{{\sqrt 5 \left( {\sqrt 3 – 1} \right)}}{{1 – \sqrt 3 }}} \right).\left( {\sqrt 7 – \sqrt 5 } \right)\\
= \left( { – \sqrt 7 – \sqrt 5 } \right)\left( {\sqrt 7 – \sqrt 5 } \right)\\
= – \left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 – \sqrt 5 } \right)\\
= – \left( {7 – 5} \right)\\
= – 2
\end{array}$