các bn ơi đoạn sau mik viết nhầm đấy bỏ phần không có ngặc đi nha

a) \(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(\Leftrightarrow A=\frac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow A=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow A=\frac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow A=\frac{-4\sqrt{x}}{\sqrt{x}-2}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)

\(\Leftrightarrow A=\frac{4x}{\sqrt{x}-3}\)

b) Để \(A=-1\)

\(\Leftrightarrow\frac{4x}{\sqrt{x}-3}=-1\)

\(\Leftrightarrow4x=3-\sqrt{x}\)

\(\Leftrightarrow4x+\sqrt{x}-3=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=-1\left(ktm\right)\\\sqrt{x}=\frac{3}{4}\Leftrightarrow x=\frac{9}{16}\left(tm\right)\end{cases}}\)

Vậy để \(A=-1\Leftrightarrow x=\frac{9}{16}\)

c) Khi \(x=36\)

\(\Leftrightarrow A=\frac{4\cdot36}{\sqrt{36}-3}=\frac{144}{3}=48\)

a) \(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right):\left(\frac{\sqrt{x}-1}{\left(x-2\sqrt{x}\right)}-\frac{2}{\sqrt{x}}\right)\)

\(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)

\(A=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(x-2\right)}\right):\left(\frac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(A=\left(\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(A=\left(\frac{-8\sqrt{x}-4x}{\left(\sqrt{x}+2\right)\sqrt{x}}\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

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

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

\(A=\frac{-4\sqrt{x}\left(2-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(3-\sqrt{x}\right)}\)

.......... Đến đây bạn tự nhân đa thức với đa thức xog rút gọn nha.

Bài 1

a, \(\left(\frac{\sqrt{y}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\sqrt{x}\left(\sqrt{y}-1\right)}{\sqrt{y}-1}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)

=\(\left(\sqrt{y}+\sqrt{x}\right).\sqrt{y}\left(\sqrt{x}-1\right)\)

b,\(\sqrt{8+2.2\sqrt{2}+1}-\sqrt{8-2.2\sqrt{2}+1}\)

=\(\sqrt{\left(\sqrt{8}+1\right)^2}-\sqrt{\left(\sqrt{8}-1\right)^2}\)

=\(\sqrt{8}+1-\left(\sqrt{8}-1\right)\)

=2

Bài 2

a, ĐKXĐ : x\(\ge\)0, x\(\pm\)1

b, Q=\(\left(\frac{\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}\right)+\frac{3-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

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

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

=\(\frac{-3}{1+\sqrt{x}}\)

c, de Q = 2 => \(\frac{-3}{1+\sqrt{x}}\)=2 =>1+\(\sqrt{x}\)=-6 =>\(\sqrt{x}\)=-7 =>x vô nghiệm