Hỏi nhiều thế.

\(=\dfrac{\sqrt{13+2\sqrt{12}}-\sqrt{13-2\sqrt{12}}+2\sqrt{12}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{12}+1-\sqrt{12}+1+2\sqrt{12}}{\sqrt{2}}\)

\(=\dfrac{2\sqrt{12}+2}{\sqrt{2}}=2\sqrt{6}+\sqrt{2}\)

Đặt \(A=\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\)

<=> \(A^2=\left(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}\right)^2\)

<=> \(A^2=6,5+\sqrt{12}+2\sqrt{\left(6,5+\sqrt{12}\right)\left(6,5-\sqrt{12}\right)}+6,5-\sqrt{12}\)

<=> \(A^2=13+2\sqrt{42,25-12}\)

<=> \(A^2=13+2\sqrt{\frac{121}{4}}\)

<=> \(A^2=13+2\cdot\frac{11}{2}=13+11=24\)

=> \(A=2\sqrt{6}\)

Vậy \(\sqrt{6,5+\sqrt{12}}+\sqrt{6,5-\sqrt{12}}+2\sqrt{6}=4\sqrt{6}\)

Lời giải:

1)

Để biểu thức có nghĩa thì:

\(2x^2-5x+3\geq 0\)

\(\Leftrightarrow 2x(x-1)-3(x-1)\geq 0\)

\(\Leftrightarrow (2x-3)(x-1)\geq 0\)

\(\Leftrightarrow \left[\begin{matrix} x\geq \frac{3}{2}\\ x\leq 1\end{matrix}\right.\)

2)

\(\sqrt{6.5+\sqrt{12}}+\sqrt{6.5-\sqrt{12}}+2\sqrt{6}\)

\(=\sqrt{(\sqrt{6})^2+(\frac{1}{\sqrt{2}})^2+2\sqrt{6}.\frac{1}{\sqrt{2}}}+\sqrt{(\sqrt{6})^2+(\frac{1}{\sqrt{2}})^2-2\sqrt{6}.\frac{1}{\sqrt{2}}}+2\sqrt{6}\)

\(=\sqrt{(\sqrt{6}+\frac{1}{\sqrt{2}})^2}+\sqrt{(\sqrt{6}-\frac{1}{\sqrt{2}})^2}+2\sqrt{6}\)

\(=\sqrt{6}+\frac{1}{\sqrt{2}}+\sqrt{6}-\frac{1}{\sqrt{2}}+2\sqrt{6}=4\sqrt{6}\)

c)

\(\sqrt{2}C=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)

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

\(=\sqrt{5}+1-\left(\sqrt{5}-1\right)-2=0\Rightarrow C=0\)

b)  

\(B=3\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)\)

\(\Rightarrow\sqrt{2}B=3\left(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\right)-\sqrt{5}\left(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\right)\)

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

\(\sqrt{2}B=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\Rightarrow B=2\sqrt{10}\)

C)√3+√5−√3−√5−√2b) (3−√5)√3+√5+(3+√5)√3−√5d) √4−√7−√4+√7+√7e) √6,5+√12+√6,5−√12+2√6mình cần giải gấp ạ 

\(A=\dfrac{5.\left(38^2-17^2\right)}{8\left(47^2-19^2\right)}\\ =\dfrac{5\left(38-17\right)\left(38+17\right)}{8\left(47-19\right)\left(47+19\right)}\\ =\dfrac{5.21.55}{8.28.66}\\ =\dfrac{5.1155}{8.1848}\\ =\dfrac{5.5}{8.8}\\ =\dfrac{25}{64}\)

\(B=\sqrt{\dfrac{0,2\times1,21\times0,3}{7,5\times3,2\times0,64}}\\ =\sqrt{0,0625\times1,890625\times0,04}\\ =\sqrt{\dfrac{121}{25600}}\\ =\dfrac{11}{160}\)

\(A=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=2\left(16-15\right)=2\)

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

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

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

\(=\frac{1}{\sqrt{2}}\left(2\sqrt{5}-2+2\sqrt{5}+2\right)=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

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

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

\(=\frac{1}{\sqrt{2}}\left(\sqrt{5}+1-\sqrt{5}+1-2\right)=0\)

\(D=\frac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}\right)\)

\(=\frac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{14}\right)\)

\(=\frac{1}{\sqrt{2}}\left(\sqrt{7}-1-\sqrt{7}-1+\sqrt{14}\right)\)

\(=\frac{1}{\sqrt{2}}\left(-2+\sqrt{14}\right)=\sqrt{7}-\sqrt{2}\)

\(E=\frac{1}{\sqrt{2}}\left(\sqrt{13+2\sqrt{12}}+\sqrt{13-2\sqrt{12}}\right)+2\sqrt{6}\)

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

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

\(=\sqrt{24}+2\sqrt{6}=4\sqrt{6}\)

1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

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

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

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

\(=4\)

2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

\(=3-\sqrt{3}+3+\sqrt{3}\)

\(=6\)

Lời giải:
a.

\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)

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

$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

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

$=|\sqrt{3}-3|+|\sqrt{3}+3|$

$=(3-\sqrt{3})+(\sqrt{3}+3)=6$

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

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

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

\(A=\sqrt{\left(9\sqrt{2}+2\sqrt{3}\right)^2}-\sqrt{\left(9\sqrt{2}-\sqrt{3}\right)^2}\)

\(=\left|9\sqrt{2}+2\sqrt{3}\right|-\left|9\sqrt{2}-\sqrt{3}\right|\)

\(=9\sqrt{2}+2\sqrt{3}-9\sqrt{2}+\sqrt{3}=3\sqrt{3}\)

Kiểm tra lại đề bài câu B, chỗ \(\sqrt{2+\sqrt{2+2}}\)

Nếu câu B sửa đề thành:

\(B=\sqrt{2+\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

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

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

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

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

\(=\sqrt{4-2}=\sqrt{2}\)