Bạn viết lại cái biểu thức được không?

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

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

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

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

2: Thay x=9 vào A, ta được:

\(A=\dfrac{3}{3+1}=\dfrac{3}{4}\)

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn.

\(A=\dfrac{\sqrt{4-2\sqrt{3}}-1}{\sqrt{4+2\sqrt{3}}+2}\)

\(A=\dfrac{\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}-1}{\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}+2}\)

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

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

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

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

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

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

\(A=\dfrac{5+3\sqrt{3}}{2}\)

A = \(\dfrac{5+3\sqrt{3}}{2}\)

a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

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

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

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

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

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

c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)

d: |B|=A

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

=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)

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

=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)

Tham khảo: https://hoc24.vn/cau-hoi/cho-adfrac12sqrtsqrt2dfrac18-dfrac18sqrt2-tinh-xa2sqrta4a21.1843666947439

Tham khảo:

Tính giá trị biểu thức A = \(x^2+\sqrt{x^{^4}+x+1}\) với x =\(\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{\sqrt{2}}{... - Hoc24

\(a=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}-\dfrac{1}{8}\sqrt{2}\\ \Leftrightarrow a+\dfrac{\sqrt{2}}{8}=\dfrac{1}{2}\sqrt{\sqrt{2}+\dfrac{1}{8}}\\ \Leftrightarrow\left(a+\dfrac{\sqrt{2}}{8}\right)^2=\dfrac{1}{4}\left(\sqrt{2}+\dfrac{1}{8}\right)\\ \Leftrightarrow a^2+\dfrac{a\sqrt{2}}{4}+\dfrac{1}{32}=\dfrac{\sqrt{2}}{4}+\dfrac{1}{32}\\ \Leftrightarrow a^2=\dfrac{\sqrt{2}-a\sqrt{2}}{4}=\dfrac{\sqrt{2}\left(1-a\right)}{4}\\ \Leftrightarrow a^4=\dfrac{a^2-2a+1}{8}\\ \Leftrightarrow a^4+a^2+1=\dfrac{a^2-2a+1}{8}+a^2+1=\dfrac{9a^2-2a+9}{8}\)

\(\Leftrightarrow a^2+\sqrt{a^4+a^2+1}=a^2+\dfrac{\sqrt{9a^2-2a+9}}{2\sqrt{2}}=\dfrac{2a^2\sqrt{2}+\sqrt{9a^2-2a+9}}{2\sqrt{2}}\)