Câu 1:

Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

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

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

Câu 3: 

Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

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

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

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

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

a) ĐKXĐ : \(x\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)

b) \(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right).\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(=\dfrac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\left(x-2\sqrt{x}+1\right)\)

\(=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)

c) Có : \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)

Khi đó B = \(\dfrac{\sqrt{3}-1}{2}-1=\dfrac{\sqrt{3}-3}{2}\)

\(a,\) B có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(b,B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)

\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)

\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}-x}{1+\sqrt{x}}\)

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

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

\(=\sqrt{x}-1\)

\(c,x=\dfrac{2-\sqrt{3}}{2}\Rightarrow B=\sqrt{\dfrac{2-\sqrt{3}}{2}}-1\)

\(=\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}.\sqrt{2}}-\sqrt{2}\) (Nhân \(\sqrt{2}\) để khử căn dưới mẫu)

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

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

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

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

a)

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

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

b)

\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)

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

=(căn 3-căn 2)(căn 3+căn 2)

=3-2=1

b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)

=(2-căn 11)(2+căn 11)

=4-11

=-7

1:

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

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)

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

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

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

\(=2x+6\sqrt{x}+1\)

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

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

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

\(=-x+8\sqrt{x}+1\)

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

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

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

\(=4x-3\sqrt{x}-3\)

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

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

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

\(=-x+3\sqrt{x}-7\)

\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)

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

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

\(=x-4-4x-6\sqrt{x}+4\)

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

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}-\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\dfrac{2}{\sqrt{2}-\sqrt{x}}-\dfrac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)

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

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

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{-\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b) Ta có: \(x=3-2\sqrt{2}\)

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

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

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

\(P=\dfrac{\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

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

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

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

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

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???

b.

\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)

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

\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)

Nguyễn Hoàng trung: Chả qua nếu $a=6-3\sqrt{3}; b=2+\sqrt{3}$ thì kết quả sẽ đẹp hơn. Còn như đề thì vẫn rút gọn được.

\(A=\frac{a-\sqrt{ab}+b}{(\sqrt{a}-\sqrt{b})^2}=\frac{a-\sqrt{ab}+b}{a-2\sqrt{ab}+b}\)

\(2a^2=12-6\sqrt{3}=(3-\sqrt{3})^2\Rightarrow a=\frac{3-\sqrt{3}}{\sqrt{2}}\) (do $a\geq 0$)

\(2b^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow b=\frac{\sqrt{3}+1}{\sqrt{2}}\) (do $b\geq 0$)

\(\Rightarrow a+b=2\sqrt{2}; ab=\frac{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}{2}=\sqrt{3}\)

Do đó: $A=\frac{2\sqrt{2}-\sqrt[4]{3}}{2\sqrt{2}-2\sqrt[4]{3}}$

\(a,\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

\(b,A=\dfrac{\sqrt{a}}{\sqrt{a}-5}-\dfrac{10\sqrt{a}}{a-25}-\dfrac{5}{\sqrt{a}+5}\)

\(\Rightarrow A=\dfrac{\sqrt{a}\left(\sqrt{a}+5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\left(\sqrt{a}-5\right)}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a+5\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{10\sqrt{a}}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}-\dfrac{5\sqrt{a}-25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{a-10\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{\left(\sqrt{a}-5\right)^2}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}\)

\(\Rightarrow A=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)

a: \(=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

b: \(A=\dfrac{a+5\sqrt{a}-10\sqrt{a}-5\sqrt{a}+25}{\left(\sqrt{a}-5\right)\left(\sqrt{a}+5\right)}=\dfrac{\left(\sqrt{a}-5\right)^2}{a-25}=\dfrac{\sqrt{a}-5}{\sqrt{a}+5}\)