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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Thực hiện phép tính: a) (1/2 √(1/2)-3/2 √4,5+2/5 √50)÷4/15 √1/8 b) √(√3-3)^2×√1/(3-√3) c) √√2 -1×√√2 +1

Toán Lớp 9: Thực hiện phép tính:
a) (1/2 √(1/2)-3/2 √4,5+2/5 √50)÷4/15 √1/8
b) √(√3-3)^2×√1/(3-√3)
c) √√2 -1×√√2 +1

Comments ( 1 )

  1. Giải đáp:
    $\begin{array}{l}
    a)\left( {\dfrac{1}{2}\sqrt {\dfrac{1}{2}}  – \dfrac{3}{2}\sqrt {4,5}  + \dfrac{2}{5}\sqrt {50} } \right):\dfrac{4}{{15}}\sqrt {\dfrac{1}{8}} \\
     = \left( {\dfrac{1}{2}.\dfrac{{\sqrt 2 }}{2} – \dfrac{3}{2}\sqrt {\dfrac{9}{2}}  + \dfrac{2}{5}.5\sqrt 2 } \right):\left( {\dfrac{4}{{15}}.\dfrac{1}{{2\sqrt 2 }}} \right)\\
     = \left( {\dfrac{{\sqrt 2 }}{4} – \dfrac{3}{2}.\dfrac{{3\sqrt 2 }}{2} + 2\sqrt 2 } \right):\dfrac{{\sqrt 2 }}{{15}}\\
     = \left( {\dfrac{{\sqrt 2 }}{4} – \dfrac{{9\sqrt 2 }}{4} + 2\sqrt 2 } \right).\dfrac{{15}}{{\sqrt 2 }}\\
     = \left( { – 2\sqrt 2  + 2\sqrt 2 } \right).\dfrac{{15}}{{\sqrt 2 }}\\
     = 0\\
    b)\sqrt {{{\left( {\sqrt 3  – 3} \right)}^2}} .\sqrt {\dfrac{1}{{3 – \sqrt 3 }}} \\
     = \left( {3 – \sqrt 3 } \right).\dfrac{1}{{\sqrt {3 – \sqrt 3 } }}\\
     = \sqrt {3 – \sqrt 3 } \\
    c)\sqrt {\sqrt 2  – 1} .\sqrt {\sqrt 2  – 1} \\
     = \sqrt {\left( {\sqrt 2  – 1} \right)\left( {\sqrt 2  + 1} \right)} \\
     = \sqrt {{2^2} – 1} \\
     = \sqrt 1 \\
     = 1
    \end{array}$

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222-9+11+12:2*14+14 = ? ( )

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