a. Không giải được\(\sqrt{29}-6\sqrt{6}< 0\)     

b. \(\left(\sqrt{8}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(2\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

a) Không thể giải vì \(\sqrt{29}-6\sqrt{6}< 0\) 

b) \(\left(\sqrt{8}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(2\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(\left(-\sqrt{2}-\sqrt{10}\right)\cdot\sqrt{2}-\sqrt{20}\) 

=\(-2-2\sqrt{5}-2\sqrt{5}\) 

=\(-2-4\sqrt{5}\) 

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

Cái này là rút gọn nha ;-; mình quên ko ghi

- Đề đầy đủ rồi nhé các bạn. KO CÓ cộng thêm căn xy bên phải đâu tại tớ nhìn bị thiếu á -.-

bạn viết lại cái đề bài đi đầy đủ ngắn gọn

Phan Văn Hiếu: Ok

Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

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

a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))

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

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

c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))

\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)

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

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

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

a) A= \(\sqrt{2-\sqrt{3}}\) \(\left(\sqrt{6}-\sqrt{2}\right)\)\(\left(2+\sqrt{3}\right)\)

A= \(\sqrt{2-\sqrt{3}}\) . \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}\) .\(\left(\sqrt{6}-\sqrt{2}\right)\)

A= \(\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\) . \(\sqrt{2+\sqrt{3}}\) . \(\sqrt{2}\left(\sqrt{3}-1\right)\)

A= 1. \(\sqrt{2\left(2+\sqrt{3}\right)}\) \(\left(\sqrt{3}-1\right)\)

A=\(\sqrt{4+2\sqrt{3}}\) .\(\left(\sqrt{3}-1\right)\)

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

A=\(\left|\sqrt{3}+1\right|\)\(\left(\sqrt{3}-1\right)\)

A=\(\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)

A=3-1

A=2

Vậy A=2

b)\(\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{\sqrt{2}+\sqrt{3}}\)=\(\frac{\sqrt{2+\sqrt{3}}.1}{\sqrt{2}+\sqrt{3}}\) = \(\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\) .

4. đặt \(\sqrt[3]{x+24}=a\) và \(\sqrt{12-x}=b\)(b>=0)

==>ta có hệ pt 

\(\int_{a^3+b^2=36}^{a+b=6}\)<=> \(\int_{a^3+\left(6-a\right)^2=36}^{b=6-a}\)<=> \(\int_{b=6-a}^{a^3+a^2-12a=0}\)<=> \(\int_{b=6-a}^{a\left(a^2+a-12\right)=0}\)<=>\(\int_{b=6-a}^{a\left(a+4\right)\left(a-3\right)=0}\)

đến đây bạn tự tìm a;b rufit hay vào tìm x là ok

3. \(\Leftrightarrow\sqrt[3]{2x^2}-\sqrt[3]{x+1}+\sqrt[3]{2x^2+1}-\sqrt[3]{x+2}=0\)

\(\Leftrightarrow\frac{2x^2-x-1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{2x^2-x-1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}=0\)

\(\Leftrightarrow2x^2-x-1=0\)

( do \(\frac{1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}>0\forall xTMĐK\))

\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2=\frac{9}{8}\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\frac{9}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{3}{4}\\x-\frac{1}{4}=-\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\) ( TM )