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

Gọi chiều dài, chiều rộng hình chữ nhật lần lượt là x ; y > 0, m 

Chu vi hình chữ nhật là : \(P=\left(a+b\right).2=46\)

Nếu tăng chiều dài 5m, giảm chiều rộng 3m thì hình chữ nhật có chiều dài gấp 4 lần chiều rộng : \(a+5=4\left(b-3\right)\)

Ta có hệ phương trình sau : \(\left\{{}\begin{matrix}\left(a+b\right).2=46\\a+5=4\left(b-3\right)\end{matrix}\right.\)

giải hệ ta được a = 15 ; b = 8 

Vậy diện tích hình chữ nhật là : \(a.b=15.8=120\)m²

a)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-4\right)+5\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\dfrac{3\sqrt{x}+1}{3}\)

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

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

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

\(a,A=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)

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

\(=\dfrac{4}{1}=4\)

Vậy \(A=4\)

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

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

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

Vậy \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) với \(x>0,x\ne1\)

a: \(=2+\sqrt{3}+2-\sqrt{3}=4\)

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

\(1,\\ a,E=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x>1\\ 2,\\ a,B=\dfrac{x-\sqrt{x}+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)\\ B=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,B=2\Leftrightarrow\sqrt{x}-1=2\left(\sqrt{x}+1\right)\\ \Leftrightarrow\sqrt{x}-1=2\sqrt{x}+2\\ \Leftrightarrow\sqrt{x}=-3\Leftrightarrow x\in\varnothing\)

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

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

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

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

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

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

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

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

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

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

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