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

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

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

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

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

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

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

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

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

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

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

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

???

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1)))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1))))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

\(A=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\left(x>0,x\ne1\right)\)

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

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

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

\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4;9\right)\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

\(C=\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

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

Điều kiện xác định \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left[\left(\sqrt[]{x}\right)^3-1\right]}{x+\sqrt[]{x}+1}-\dfrac{\sqrt[]{x}\left(2\sqrt[]{x}+1\right)}{\sqrt[]{x}}+\dfrac{2\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+2\right)}{\sqrt[]{x}-1}\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-1\right)\left(x+\sqrt[]{x}+1\right)}{x+\sqrt[]{x}+1}-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=\sqrt[]{x}\left(\sqrt[]{x}-1\right)-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=x-\sqrt[]{x}-2\sqrt[]{x}-1+2\sqrt[]{x}+4\)

\(\Rightarrow P=x-\sqrt[]{x}+3\)

b) \(A=\dfrac{P}{2012\sqrt[]{x}}=\dfrac{x-\sqrt[]{x}+3}{2012\sqrt[]{x}}\)\(\)

\(=\dfrac{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+3}{2012\sqrt[]{x}}\)

\(=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}{2012\sqrt[]{x}}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{\dfrac{11}{4}}{2012\sqrt[]{x}}=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\)

Ta lại có  \(\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}\ge0,\forall x\ne0\)

\(\dfrac{1}{\sqrt[]{x}}>0\Rightarrow\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{4.2012}=\dfrac{11}{8048}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{8048}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt[]{x}=1\Leftrightarrow x=1\)

Vậy \(GTNN\left(A\right)=\dfrac{11}{8048}\left(tạix=1\right)\)

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

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

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

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

b) \(\dfrac{P}{2012\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{2012\sqrt{x}}=\dfrac{\sqrt{x}}{2012}-\dfrac{1}{2012}+\dfrac{3}{2012\sqrt{x}}\)

\(=\left(\dfrac{\sqrt{x}}{2012}+\dfrac{3}{2012\sqrt{x}}\right)-\dfrac{1}{2012}\)

\(\ge2\sqrt{\dfrac{\sqrt{x}.3}{2012^2\sqrt{x}}}-\dfrac{1}{2012}\) (BĐT Cauchy)

\(=\dfrac{2\sqrt{3}}{2012}-\dfrac{1}{2012}=\dfrac{2\sqrt{3}-1}{2012}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{x}}{2012}=\dfrac{3}{2012\sqrt{x}}\Leftrightarrow x=3\)(tm)