Ta có \(\left|2x-1\right|=\dfrac{1}{2}\)

\(\left|2x-1\right|=2x-1\) khi \(2x-1\ge0\Leftrightarrow2x\ge1\Leftrightarrow x\ge\dfrac{1}{2}\)

\(\left|2x-1\right|=-\left(2x-1\right)\) khi \(2x-1< 0\Leftrightarrow2x< 1\Leftrightarrow x< \dfrac{1}{2}\)

Ta giải hai phương trình sau:

pt₁: \(2x-1=\dfrac{1}{2}\left(ĐK:x\ge\dfrac{1}{2}\right)\)

\(\Leftrightarrow2x=\dfrac{1}{2}-1\)

\(\Leftrightarrow2x=-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{-\dfrac{1}{2}}{2}=-\dfrac{1}{2}\left(ktm\right)\)

pt₂: \(-\left(2x-1\right)=\dfrac{1}{2}\left(ĐK:x< \dfrac{1}{2}\right)\)

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

\(\Leftrightarrow-2x=\dfrac{1}{2}-1\)

\(\Leftrightarrow-2x=-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{-\dfrac{1}{2}}{-2}=\dfrac{1}{4}\left(tm\right)\)

Vậy giá trị của đa thức \(E\left(x\right)\) tại \(x=\dfrac{1}{4}\)

\(E\left(x\right)=-4x^4+x+1=-4.\left(\dfrac{1}{4}\right)^4+\dfrac{1}{4}+1=\dfrac{79}{64}\)

Bạn vào cái ô đầu tiên trên thanh công thức để gõ lại biểu thức đi bạn. Khó nhìn quá

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

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

\(=\left(\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}}{x-1}\)

\(=\dfrac{4x^2}{\left(x-1\right)^2}\)

b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-2x^2+4x-2=0\)

\(\Leftrightarrow2x^2+4x-2=0\)

\(\Leftrightarrow x^2+2x-1=0\)

\(\Leftrightarrow\left(x+1\right)^2=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}-1\\x=\sqrt{2}-1\end{matrix}\right.\)

c: Ta có: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=2\)

Thay x=2 vào E, ta được:

\(E=\dfrac{4\cdot2^2}{1}=16\)

Với \(x\ge0\)

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

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

\(=\dfrac{4\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}\)

a) Ta có: \(E=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

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

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

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

b) Để \(E=\dfrac{8}{9}\) thì \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}-36\sqrt{x}+24=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow24x-12\sqrt{x}-48\sqrt{x}+24=0\)

\(\Leftrightarrow12\sqrt{x}\left(2\sqrt{x}-1\right)-24\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(12\sqrt{x}-24\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=0\\12\sqrt{x}-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}=1\\12\sqrt{x}=24\end{matrix}\right.\)

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

Chọn C