\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

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

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

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

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

a: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-3\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)

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

b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

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

\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)

\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)

Lời giải:

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{16}-2}-\frac{12}{3-\sqrt{16}}\right).(\sqrt{6}+11)=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4}{4-2}-\frac{12}{3-4}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{6-1}+2+12\right)(\sqrt{6}+11)=(3\sqrt{6}-3+14)(\sqrt{6}+11)\)

\(=(3\sqrt{6}+11)(\sqrt{6}+11)\)

^{Bạn tham khảo ctrl ạ!}

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)

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

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

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

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

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

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

=-2

4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

=4

a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)

\(=127-22\sqrt{6}\)

b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

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

=-1+5

=4

1

Bài 1:

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)

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

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

Bài 2:

a: ĐKXĐ: x>=3

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

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x-3\right)^2}=12\)

=>\(\left|x-3\right|=12\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

Bài 3:

a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)

\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)

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

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

b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)

\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)

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

d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)

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

\(=\dfrac{2\left(a-2\right)}{a+2}\)

ưu tiên phương pháp bình phương :

a) \(\left(4+\sqrt{15}\right)^2\left(\sqrt{10}-\sqrt{6}\right)^2\left(\sqrt{4-\sqrt{15}}\right)^2\)

\(=\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)^2\)

Tính ra kết quả nhớ căn đó

b) Phương pháp trục căn thức :

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

Trên tử có hàng đẳng thức . bạn tự quy động là ra 

mình vẫn chưa hiểu câu a

a)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

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

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

=\(\sqrt{\left(\sqrt{5}\right)^2+2.\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{16-15}\)

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

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

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

= \(5-3=2\)

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