\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)

\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)

\(A=\frac{\sqrt{x}-2}{x-4}\)

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

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

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

b: P=1/4

=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)

=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)

=>\(4\sqrt{x}-8-3\sqrt{x}=0\)

=>\(\sqrt{x}=8\)

=>x=64

c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

Ai giải giúp mình bài 1 với bài 4 trước đi

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

ĐK : x>0, x khác 1

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

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

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

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

đkxđ: \(x\ge0;x\ne4\)

\(Q=\left[\frac{x-\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-2}\right]\div\left[\frac{\sqrt{x}+2}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

\(Q=\left[\frac{x-\sqrt{x}+7+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\div\left[\frac{\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-2\right)^2-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\)

\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\div\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(Q=\frac{x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{6\sqrt{x}}\)

\(Q=\frac{\left(x+9\right)\sqrt{x}}{6x}\)

\(Q=\frac{x\sqrt{x}+9\sqrt{x}}{6x}\)

đkxđ sửa tí thành \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)

ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)

  <=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)

<=> \(A^3=x^3-3x+3A\)

<=> \(A^3-3A-x^3+3x=0\)

<=>\(\left(A^3-x^3\right)-3A+3x=0\)

<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)

<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)

<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )

vậy \(A=x\)