\(abc\le\left(\frac{a+b+c}{3}\right)^3\Leftrightarrow\sqrt[3]{abc}\le\frac{a+b+c}{3}\)

BĐT Cô- si 

đánh giá từ tbn sang tbc đấy bạn 

\(A=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)

\(A=\frac{2x+\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}-2\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}\)

\(A=2\sqrt{x}+1\)

\(\frac{1}{2+\sqrt{5}+2\sqrt{2}+\sqrt{10}}=\frac{1}{2+\sqrt{5}+\sqrt{2}\left(2+\sqrt{5}\right)}=\frac{1}{\left(\sqrt{2}+1\right)\left(2+\sqrt{5}\right)}\)

\(=\frac{\left(\sqrt{2}-1\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\sqrt{10}-2\sqrt{2}-\sqrt{5}+2\)

\(\left(\frac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)

\(\frac{x+\sqrt{x}-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\sqrt{x}-1\)

\(\frac{x+\sqrt{x}-1-x-\sqrt{x}+\sqrt{x}+1}{x+\sqrt{x}+1}\)

\(\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)

\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{\sqrt{x}-1}{1}\)

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(5\sqrt{34}+\left|6-\sqrt{34}\right|\)

\(6>\sqrt{34}\)

\(5\sqrt{34}+6-\sqrt{34}\)

\(4\sqrt{34}+6\)

=) ko bt

Đk: \(x\ne\)2; x \(\ne\)-4; -4 \(\le\)x \(\le\)2

Đặt \(\sqrt{\frac{2-x}{x+4}}=a\) (đk: \(a\ge\)0) => \(\sqrt{\frac{x+4}{2-x}}=\frac{1}{a}\)

Do đó, ta có: \(a-\frac{2}{a}+1=0\)

=> a² + a - 2 = 0

<=> a² + 2a - a - 2 = 0

<=> (a + 2)(a - 1) = 0

<=> \(\orbr{\begin{cases}a=-2\left(loại\right)\\a=1\end{cases}}\)

<=> \(\sqrt{\frac{2-x}{x+4}}=1\)

<=> \(2-x=x+4\)

<=> \(2x=-2\) <=> x = -1 (tm)

Vậy S = {-1}

ĐK: \(x\ge3\)

\(x^2+4x-5=a,x-3=b\)(\(a,b\ge0\))

Phương trình tương đương với: 

\(3\sqrt{a}+\sqrt{b}=\sqrt{11a-19b}\)

\(\Leftrightarrow9a+6\sqrt{ab}+b=11a-19b\)

\(\Leftrightarrow a-3\sqrt{ab}-10b=0\)

Với \(b=0\Rightarrow x=3\)không là nghiệm của phương trình ban đầu.

Với \(b\ne0\):

\(\frac{a}{b}-3\sqrt{\frac{a}{b}}-10=0\Leftrightarrow\sqrt{\frac{a}{b}}=5\Leftrightarrow\frac{a}{b}=25\)

\(\Rightarrow x^2+4x-5=25\left(x-3\right)\)

\(\Leftrightarrow x=\frac{1}{2}\left(21\pm\sqrt{161}\right)\)(thỏa)