\(\frac{\sqrt{7}+5}{\sqrt{7}-5}+\frac{\sqrt{7}-5}{\sqrt{7}+5}\)

\(=\frac{\left(\sqrt{7}+5\right)^2}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}+\frac{\left(\sqrt{7}-5\right)^2}{\left(\sqrt{7}+5\right)\left(\sqrt{7}-5\right)}\)

\(=\frac{32+10\sqrt{7}}{-18}+\frac{32-10\sqrt{7}}{-18}\)

\(=\frac{64}{-18}=-\frac{32}{9}\)

\(\frac{\sqrt{7}+5}{\sqrt{7}-5}+\frac{\sqrt{7}-5}{\sqrt{7}+5}\)

\(=\frac{\left(\sqrt{7}+5\right)^2+\left(\sqrt{7}-5\right)^2}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}\)

\(=\frac{7+10\sqrt{7}+25+7-10\sqrt{7}+25}{7-25}\)

\(=\frac{64}{-18}=\frac{-32}{9}\)

I don't now

mik ko biết 

sorry 

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I don't now

mik ko biết 

sorry 

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Tìm GTNN của A = \(\frac{3x^2-8x+6}{x^2-2x+1}\)

I don't now

mik ko biết 

sorry 

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1.  \(2ab\le\frac{\left(a+b\right)^2}{2}\le a^2+b^2\) (  \(\forall a;b\))

2.  \(\frac{a}{b}+\frac{b}{a}\ge2\)( \(\forall a;b>0\))

3.  \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)\(\left(a;b>0\right)\)

4.  \(\frac{1}{ab}\ge\frac{4}{\left(a+b\right)^2}\) \(\left(a;b>0\right)\)

5.  \(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

6.  \(a^2+b^2+c^2\ge ab+bc+ca\)

7.  \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

8.  \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) \(\left(a;b;c>0\right)\)

9.  \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)\(\left(x;y>0\right)\)

10.  \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) \(\left(x;y;z>0\right)\)

I don't now

mik ko biết 

sorry 

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i don;t no

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vui mà

vui mak, bạn dis bài mk trg khi bài mk đúng, vui quá mak

cái loại hạ đẳng thì chỉ có vậy thôi, mak thôi cx đúng súc vật đc z là hiếm lắm oy

Ta có \(\left(\sqrt{x}-1\right)^2\ge0\)

=> \(x-2\sqrt{x}+1\ge0\)=> \(x-\sqrt{x}+1\ge\sqrt{x}\)Mà trên kia căn x lại chia hết

=> 2 vế bằng nhau => x = 1