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

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

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

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

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

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

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

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

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

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

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

\(6>\sqrt{34}\)

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

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

=) 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) 

A = \(\frac{\sqrt{11-6\sqrt{2}}+\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}}{\sqrt{2+\sqrt{3}}+\sqrt{14-5\sqrt{3}}}\)

A = \(\frac{\sqrt{2\left(9-6\sqrt{2}+2\right)}+\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{4+2\sqrt{3}}+\sqrt{28-10\sqrt{3}}}\)

A = \(\frac{\sqrt{2\left(3-\sqrt{2}\right)^2}+\sqrt{5+2\sqrt{5}+1}+\sqrt{9-6\sqrt{5}+5}-2}{\sqrt{3+2\sqrt{3}+1}+\sqrt{25-10\sqrt{3}+3}}\)

A = \(\frac{\sqrt{2}\left(3-\sqrt{2}\right)+\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}-2}{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(5-\sqrt{3}\right)^2}}\)

A = \(\frac{3\sqrt{2}-2+\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{3}+1+5-\sqrt{3}}\)

A = \(\frac{3\sqrt{2}}{6}=\frac{1}{\sqrt{2}}\)

Đáp án:

a) góc ACD = 60o60o

b) CD=3+3√3

Giải thích các bước giải:

a) Vì AB=OA=OB nên tam giác OAB là tam giác đều

⇒ góc OAB=góc OBA= 60o60o

⇒ góc OBC=180o180o -60o60o=120o120o

Xét tam giác OBC có OC=AB=OB ⇒ tam giác OBC cân tại B

⇒ góc BOC= góc BCO

Mà góc BOC+góc BCO=180o180o -120o120o=60o60o

⇒ góc BCO hay góc ACD bằng 60o60o

b) Kẻ OH ⊥AB

ta có: OH= 3√323√32

HC=HB+BC= 3232 +3=9292

⇒ OC= 2√OH2+HC2OH2+HC22 =3√3

⇒ CD=CO+OC=3+3√3

tại sao OA = AB = OB