1, Với \(x\ge0,x\ne1\) ta có :

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

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

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

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

2, Ta có \(P=\dfrac{7}{4}\)

          \(\Rightarrow\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{7}{4}\)

         \(\Leftrightarrow4\left(2\sqrt{x}+1\right)=7\left(\sqrt{x}+1\right)\)

         \(\Leftrightarrow8\sqrt{x}+4=7\sqrt{x}=7\)

          \(\Leftrightarrow\sqrt{x}=3\)

          \(\Leftrightarrow x=9\left(tm\right)\)

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

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

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

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

2) Để \(P=\dfrac{7}{4}\) thì \(\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}=\dfrac{7}{4}\)

\(\Leftrightarrow4\cdot\left(2\sqrt{x}+1\right)=7\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow8\sqrt{x}+4=7\sqrt{x}+7\)

\(\Leftrightarrow8\sqrt{x}-7\sqrt{x}=7-4\)

\(\Leftrightarrow\sqrt{x}=3\)

hay x=9(nhận)

Vậy: Để \(P=\dfrac{7}{4}\) thì x=9

Ai lm đc câu nào thì giúp mk với , cảm ơn !!

\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)

a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)

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

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

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

Để \(P=\dfrac{7}{2}\) thì \(2x+2\sqrt{x}+2-7\sqrt{x}=0\)

\(\Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{4}\end{matrix}\right.\)

a: |x|+2003>=2003

=>A<=2022/2003

Dấu = xảy ra khi x=0

b: |x|+1>=1

=>(|x|+1)^10>=1

=>B>=2010

Dấu = xảy ra khi x=0

\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)

\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{20a-11}{2012}\\x_1x_2=-1\end{matrix}\right.\)

\(P=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(\dfrac{x_1-x_2}{2}-\dfrac{x_1-x_2}{x_1x_2}\right)^2\)

\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}-\dfrac{1}{x_1x_2}\right)^2\)

\(=\dfrac{3}{2}\left(x_1-x_2\right)^2+2\left(x_1-x_2\right)^2\left(\dfrac{1}{2}+1\right)^2\)

\(=6\left(x_1-x_2\right)^2=6\left(x_1+x_2\right)^2-24x_1x_2\)

\(=6\left(\dfrac{20a-11}{2012}\right)^2+24\ge24\)

Dấu "=" xảy ra khi \(a=\dfrac{11}{20}\)

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

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)