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

Toán Lớp 9: a) 5+√5/5-√5 + 5-√5/5+√5 b) 3/√5+√2 + 1/√2-1 – 4/3-√5 c) √5-2/5+2√5 – 1/2+√5 + 1/√5 Mọi người giải giúp em

Toán Lớp 9: a) 5+√5/5-√5 + 5-√5/5+√5
b) 3/√5+√2 + 1/√2-1 – 4/3-√5
c) √5-2/5+2√5 – 1/2+√5 + 1/√5
Mọi người giải giúp em

Comments ( 1 )

  1. Lời giải và giải thích chi tiết:
    $a)\frac{5+\sqrt5}{5-\sqrt5}+\frac{5-\sqrt5}{5+\sqrt5}$
    $=\frac{\sqrt5(\sqrt5+1)}{\sqrt5(\sqrt5-1)}+\frac{\sqrt5(\sqrt5-1)}{\sqrt5(\sqrt5+1)}$
    $=\frac{(\sqrt5+1)^2}{(\sqrt5-1)(\sqrt5+1)}+\frac{(\sqrt5-1)^2}{(\sqrt5-1)(\sqrt5+1)}$
    $=\frac{6+2\sqrt5+6-2\sqrt5}{(\sqrt5-1)(\sqrt5+1)}$
    $=\frac{12}{5-1}$
    $=\frac{12}{4}=3$
    $b)\frac{3}{\sqrt5+\sqrt2}+\frac{1}{\sqrt2-1}-\frac{4}{3-\sqrt5}$
    $=\frac{3(\sqrt2-1)(3-\sqrt5)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}+\frac{(\sqrt5+\sqrt2)(3-\sqrt5)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}-\frac{4(\sqrt5+\sqrt2)(\sqrt2-1)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$
    $=\frac{9\sqrt2-3\sqrt{10}+3\sqrt5-9}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}+\frac{3\sqrt5-5+3\sqrt2-\sqrt{10}}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}-\frac{4\sqrt{10}-4\sqrt5+8-4\sqrt2}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$
    $=\frac{9\sqrt2-3\sqrt{10}+3\sqrt5-9+3\sqrt5-5+3\sqrt2-\sqrt{10}-(4\sqrt{10}-4\sqrt5+8-4\sqrt2)}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$
    $=\frac{9\sqrt2-3\sqrt{10}+3\sqrt5-9+3\sqrt5-5+3\sqrt2-\sqrt{10}-4\sqrt{10}+4\sqrt5-8+4\sqrt2}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$
    $=\frac{16\sqrt2-8\sqrt{10}+10\sqrt5-22}{(\sqrt5+\sqrt2)(\sqrt2-1)(3-\sqrt5)}$
    Kho’k tiếng chos ;-; ;-; ;-;
    $c)\frac{\sqrt5-2}{5+2\sqrt5}-\frac{1}{2+\sqrt5}+\frac1{\sqrt5}$
    $=\frac{\sqrt5-2}{\sqrt5(\sqrt5+2)}-\frac{\sqrt5}{\sqrt5(\sqrt5+2)}+\frac{\sqrt5+2}{\sqrt5(\sqrt5+2)}$
    $=\frac{\sqrt5-2-\sqrt5+\sqrt5+2}{\sqrt5(\sqrt5+2)}$
    $=\frac{\sqrt5}{\sqrt5(\sqrt5+2)}$
    $=\frac{1}{\sqrt5+2}$

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

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