\(\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\sqrt{2}\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{3-2\sqrt{3}+1}}}=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{\sqrt{2}.\sqrt{4-\sqrt{3}}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)

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

\(a.\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}=\sqrt{x}-\sqrt{y}+\sqrt{x}+\sqrt{y}=2\sqrt{x}\)

\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)

\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)

Câu a

\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\sqrt{x}+\sqrt{y}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\\ =\dfrac{2x\sqrt{y}}{\sqrt{xy}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)

Bạn nhóm hai thừa số đầu, hai thừa số cuối, sau đó áp dụng hằng đẳng thức số 3 ta dc

A=[(√2+√3)^2-5][5-(√2-√3)^2]

= Bạn khai triển các bình phương ra, thu gọn, kết quả là 24

Bạn tách các bt trong ngoặc ra về các hằng đẳng thức số1, 2 như này

8-2√15= 3-2.√3.√5+5=

(√3+√5)^2.

Nhưng căn tiếp theo phải là 21-4√5 thì mới ra được là (2√5-1)^2

Những câu tiếp theo hoàn toàn tương tự

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

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

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

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

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)

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

1: ĐKXĐ: x+3>=0

=>x>=-3

\(\sqrt{x+3}>2\)

=>x+3>4

=>x>4-3=1

2: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 1\)

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-1< 0\)

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

=>\(\dfrac{3}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

3: ĐKXĐ: x>=0

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

=>\(x-4\sqrt{x}+3-5=x+2\sqrt{x}-5\)

=>\(x-4\sqrt{x}-2-x-2\sqrt{x}+5=0\)

=>\(-6\sqrt{x}+3=0\)

=>\(-6\sqrt{x}=-3\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>x=1/4(nhận)