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

Toán Lớp 9: Rút gọn : Bằng cách bình phương 2 vế H = √((3 + √2)/(3- √2)) – √((3-2√2) / ( 3 + 2√2))

Toán Lớp 9: Rút gọn : Bằng cách bình phương 2 vế
H = √((3 + √2)/(3- √2)) – √((3-2√2) / ( 3 + 2√2))

Comments ( 1 )

  1. \text{Giải đáp:}
    \sqrt{{26-2\sqrt{7})/(5+3\sqrt{2}}
    \text{Lời giải và giải thích chi tiết:}
    H=\sqrt{(3+\sqrt{2})/(3-\sqrt{2})}-\sqrt{(3-2\sqrt{2})/(3+2\sqrt{2})}
    H^2=(\sqrt{(3+\sqrt{2})/(3-\sqrt{2})}-\sqrt{(3-2\sqrt{2})/(3+2\sqrt{2})})^2
    H^2=(\sqrt{(3+\sqrt{2})/(3-\sqrt{2})})^2-2\sqrt{(3+\sqrt{2})/(3-\sqrt{2})}*\sqrt{(3-2\sqrt{2})/(3+2\sqrt{2})}+(\sqrt{(3-2\sqrt{2})/(3+2\sqrt{2})})^2
    H^2=(3+\sqrt{2})/(3-\sqrt{2})-2\sqrt{(3+\sqrt{2})/(3-\sqrt{2})*(3-2\sqrt{2})/(3+2\sqrt{2})}+(3-2\sqrt{2})/(3+2\sqrt{2})
    H^2=((3+\sqrt{2})/(3-\sqrt{2})+(3-2\sqrt{2})/(3+2\sqrt{2}))-2\sqrt{(3+\sqrt{2})/(3-\sqrt{2})*(3-2\sqrt{2})/(3+2\sqrt{2})}
    H^2=((3+\sqrt{2})(3+2\sqrt{2})+(3-2\sqrt{2})(3-\sqrt{2}))/((3-\sqrt{2})(3+2\sqrt{2}))-2\sqrt{((3+\sqrt{2})(3-2\sqrt{2}))/((3-\sqrt{2})(3+2\sqrt{2})
    H^2=(3^2+6\sqrt{2}+3\sqrt{2}+2(\sqrt{2})^2+3^2-3\sqrt{2}-6\sqrt{2}+2(\sqrt{2})^2)/(3^2+ 6\sqrt{2}-3\sqrt{2}-2(\sqrt{2})^2)-2*\sqrt{3^2-6\sqrt{2}+3\sqrt{2}-2(\sqrt{2})^2}/\sqrt{3^2+6\sqrt{2} -3\sqrt{2}\-2(\sqrt{2} ) ^2
    H^2=(26)/(5+3\sqrt{2})-(2\sqrt{5-3\sqrt{2}))/(\sqrt{5+3\sqrt{2})
    H^2=(26-2\sqrt{5-3\sqrt{2})*\sqrt{5+3\sqrt{2}))/(5+3\sqrt{2})
    H^2=(26-2\sqrt{5^2-(3\sqrt{2})^2))/(5+3\sqrt{2})
    H^2=(26-2\sqrt{25-18))/(5+3\sqrt{2})
    H^2=(26-2\sqrt{7})/(5+3\sqrt{2})
    => H=\sqrt{{26-2\sqrt{7})/(5+3\sqrt{2}}

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