mình trình bày trong hình, hơi mờ bạn thông cảm!

như này là xịn lém rrr. Thănkiu nhìu nhó Nguyen Thu Hong

Lời giải:

Gọi biểu thức cần rút gọn là $P$

Xét tử số: $\sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{3+2\sqrt{3.1}+1}-\sqrt{3}$

$=\sqrt{(\sqrt{3}+1)^2}-\sqrt{3}=|\sqrt{3}+1|-\sqrt{3}=1$

Xét mẫu số:

Ta dự đoán sẽ rút gọn được $\sqrt[3]{17\sqrt{5}-38}$

Đặt $17\sqrt{5}-38=(a+\sqrt{5})^3$ với $a$ nguyên.
$\Leftrightarrow 17\sqrt{5}-38=a^3+15a+\sqrt{5}(3a^2+5)$

$\Rightarrow 17=3a^2+5$ và $-38=a^3+15a$

$\Rightarrow a=-2$

Vậy $17\sqrt{5}-38=(-2+\sqrt{5})^3$

$\Rightarrow (\sqrt{5}+2)\sqrt[3]{17\sqrt{5}-38}=(\sqrt{5}+2)(-2+\sqrt{5})=1$

Vậy $P=\frac{1}{1}=1$

Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)

\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)

\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)

\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)

\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)

\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)

\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)

\(=60\)

Bài 1:

\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)

Bài 2:

\(a,P=\left(\frac{1}{x+\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+2\sqrt{x}+1}\left(x>0\right)\)

\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\times\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)

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

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

\(=\frac{1-x}{x}\)

\(b,\forall x>0\Leftrightarrow\frac{1-x}{x}>\frac{1}{2}\)

\(\Leftrightarrow2\left(1-x\right)>x\)

\(\Leftrightarrow2-2x>x\)

\(\Leftrightarrow-3x>-2\)

\(\Leftrightarrow x< \frac{2}{3}\)

\(\Rightarrow P>\frac{1}{2}\Leftrightarrow\forall0< x< \frac{2}{3}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)

\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

\(=\sqrt{6-2+2\sqrt{3}}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(1+\sqrt{3}\right)^2}=1+\sqrt{3}\)