a: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)

\(=\sqrt{4-2\cdot2\cdot\sqrt2+2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)

\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{\left(3\sqrt2-2\right)^2}\)

\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)

b: \(\sqrt{\left(\sqrt3-\sqrt2\right)^2}+\sqrt2=\sqrt3-\sqrt2+\sqrt2=\sqrt3\)

c: \(3\sqrt5-\sqrt{\left(1-\sqrt5\right)^2}\)

\(=3\sqrt5-\left|1-\sqrt5\right|\)

\(=3\sqrt5-\left(\sqrt5-1\right)=2\sqrt5+1\)

d:Sửa đề: \(\sqrt{17-12\sqrt2}+\sqrt{6+4\sqrt2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt2+8}+\sqrt{4+2\cdot2\cdot\sqrt2+2}\)

\(=\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{\left(2+\sqrt2\right)^2}=3-2\sqrt2+2+\sqrt2=5-\sqrt2\)

VO HAN

bạn nào giải ra bằng 1 thì mới đúng

1/ \(A=\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}-\sqrt{3}>0\))

\(B=\sqrt{6+2\sqrt{5}}-\sqrt{13}+\sqrt{48}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{13}+4\sqrt{3}=\left|\sqrt{5}+1\right|-\sqrt{13}+4\sqrt{3}=\sqrt{5}+1+\sqrt{13}+4\sqrt{5}\)

2/Ta có :

\(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{3\sqrt{2}}{3\sqrt{3}-3}-\frac{5\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{3\sqrt{2}}{3\left(\sqrt{3}-1\right)}-\frac{5\sqrt{6}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}-1\right)}\right).\frac{1}{\sqrt{6}}\)

\(=\frac{3\sqrt{2}-15\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-12\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-7+\sqrt{3}}{6}\)

Vậy...

Bài 1:

Ta có: \(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}-2\cdot\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)

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

\(=\left|\sqrt{5}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|-2\cdot\left|\sqrt{5}-1\right|\)

\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)

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

=2

Vậy: A=2

Bài 2: Sửa đề: Chứng minh \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}=\frac{-7+\sqrt{3}}{6}\)

Ta có: \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{9\sqrt{2}}{3\left(\sqrt{27}-3\right)}-\frac{\sqrt{150}\left(\sqrt{27}-3\right)}{3\cdot\left(\sqrt{27}-3\right)}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{9\sqrt{2}-45\sqrt{2}+3\sqrt{150}}{9\left(\sqrt{3}-1\right)}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-36\sqrt{2}+3\sqrt{150}}{9\sqrt{6}\cdot\left(\sqrt{3}-1\right)}\)

\(=\frac{\sqrt{54}\cdot\left(5-4\sqrt{3}\right)}{\sqrt{486}\cdot\left(\sqrt{3}-1\right)}\)

\(=\frac{5-4\sqrt{3}}{3\sqrt{3}-3}\)

\(=\frac{-7+\sqrt{3}}{6}\)(đpcm)

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

=2.

\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)

\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)

Giup mình phần 3,4,5 của bài 2 với bài 4 nữa . Helpppp me !!

1/ \(\sqrt{x-2}-\sqrt{1-3x}=0\\ đk:\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

=> pt vô no

2/ \(\sqrt{15-x}+\sqrt{3-x}=6\\ đk\left\{{}\begin{matrix}15-x\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le15\\x\le3\end{matrix}\right.\Leftrightarrow x\le3\)

\(pt\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow2\sqrt{\left(15-x\right)\left(3-x\right)}=2x+36\)

\(\Leftrightarrow4\left(15-x\right)\left(3-x\right)=\left(2x+18\right)^2\left(đk:x\ge-9\right)\)

\(\Leftrightarrow-144x=144\Leftrightarrow x=-1\left(nhan\right)\)

Câu 1: ĐKXĐ: \(\left\{{}\begin{matrix}x-2\ge0\\1-3x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le\frac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ \(\Rightarrow\) pt vô nghiệm

Câu 2:

ĐKXĐ: \(x\le3\)

\(\Leftrightarrow15-x+3-x+2\sqrt{\left(15-x\right)\left(3-x\right)}=36\)

\(\Leftrightarrow x+9=\sqrt{x^2-18x+45}\) (\(x\ge-9\))

\(\Leftrightarrow x^2+18x+81=x^2-18x+45\)

\(\Leftrightarrow36x=-36\Rightarrow x=-1\)

Câu 3:

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt{x-1}=2+\sqrt{x+1}\)

\(\Leftrightarrow x-1=4+x+1+4\sqrt{x+1}\)

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

Phương trình vô nghiệm