\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=2005\left(2x-1\right)+\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\Leftrightarrow\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{3\cdot5}}=2005\left(2x-1\right)+\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=2005\left(2x-1\right)+\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\Leftrightarrow2005\left(2x-1\right)=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Hải Ngọc nhầm 2006 thành 2005 rồi

mình đang cần gấp làm nhanh nha mọi người

2: \(=\sqrt{2}-1-\sqrt{2}=-1\)

3: \(=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\)

\(=\dfrac{7+4\sqrt{3}-7+4\sqrt{3}}{1}=8\sqrt{3}\)

4: \(=1+\dfrac{2-\sqrt{3}}{2-\sqrt{3}}=1+1=2\)

Lời giải:

a)

\(\frac{2A}{\sqrt{2}}=\frac{4+2\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{3+1+2\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3}}}+\frac{3+1-2\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3}}}\)

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

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

$\Rightarrow A=\sqrt{2}$

b)

\(B=\sqrt{10+2\sqrt{15}-2\sqrt{6}-2\sqrt{10}}=\sqrt{(8+2\sqrt{15})+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)

\(=\sqrt{(\sqrt{3}+\sqrt{5})^2+2-2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)

\(=\sqrt{(\sqrt{3}+\sqrt{5}-\sqrt{2})^2}=\sqrt{3}+\sqrt{5}-\sqrt{2}\)

c)

\(C=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}=\frac{\sqrt{(x-1)-2\sqrt{x-1}+1}+\sqrt{(x-1)+2\sqrt{x-1}+1}}{\sqrt{(x-2)^2}}\)

\(=\frac{\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+1)^2}}{|x-2|}=\frac{|\sqrt{x-1}-1|+|\sqrt{x-1}+1|}{|x-2|}\)

ko hiểu

bài này là lớp 9

minh cung khong biet !!!!!!!!!!!!!!1

Tham khảo:

1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24

\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)

\(\Leftrightarrow5x+3=4+x^2+3x\)

\(\Leftrightarrow x^2-2x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

\(a,\left(2\sqrt{3}-\sqrt{2}\right)^2+2\sqrt{24}=\left[\left(2\sqrt{3}\right)^2-2.2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\right]+2\sqrt{24}\\ =\left[12-4\sqrt{6}+2\right]+2\sqrt{24}=14-4\sqrt{6}+4\sqrt{6}=14\\ b,\left(3\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+2\sqrt{3}\right)-\sqrt{60}=3\sqrt{5}.\sqrt{5}-2\sqrt{3}.\sqrt{3}+3\sqrt{5}.2\sqrt{3}-\sqrt{3}.\sqrt{5}-\sqrt{60}\\ =15-6+6\sqrt{15}-\sqrt{15}-\sqrt{2^2.15}\\ =9+3\sqrt{15}\)