a) \(P=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\left(a>0\right)\)

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

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

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

\(\ge-\dfrac{1}{4}\)

\(\Rightarrow P_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)

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

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

\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x\sqrt{x}-2x+8\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne3\\x\ge1\end{matrix}\right.\)

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

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

b, Thấy : \(\sqrt{x-1}\ge0\)

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

Vậy \(Min=\sqrt{2}\Leftrightarrow x=1\)

Vậy ,...
 

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

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

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

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

b)\(P=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)

Vậy \(P_{min}=-\dfrac{1}{4}\)

Này, mình nói bạn Thịnh là bài này mk nghĩ ý b bạn vẫn làm được mà bạn chỉ làm mỗi ý a là sao? Làm ý a bỏ ý b hả, zì kì thế

a) \(C=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a>0.a\ne1\right)\)

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

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

b) \(C=\dfrac{1}{4}\Rightarrow-\dfrac{1}{\sqrt{a}}=\dfrac{1}{4}\Rightarrow\sqrt{a}=-4\) (vô lý) \(\Rightarrow\) không có a thỏa đề

ĐK:\(x\ge0;x\ne9\)

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

\(=\dfrac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+x-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)

b)\(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=1+\dfrac{2}{\sqrt{x}+2}\le1+\dfrac{2}{0+2}=2\)

Dấu "=" xảy ra khi \(x=0\)

Vậy \(P_{max}=2\)

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

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

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

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

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

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

\(=\dfrac{a+1}{a+\sqrt{a}+1}\)

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