a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

b) Thay x=3+2\(\sqrt{2}\)

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

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

c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0

\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)

a) \(\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right):\left(\frac{1-x}{x+1}\right)\)

= \(\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{x^2-2x-3x+6}\right):\left(\frac{1-x}{x+1}\right)\)

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

= \(\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right).\frac{x+1}{1-x}\)

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

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

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

b) Để A >1 \(\Leftrightarrow\frac{x+1}{\left(x-2\right)\left(1-x\right)}>1\)

\(\Leftrightarrow\frac{-\left(1-x\right)\left(3-x\right)}{\left(x-2\right)\left(1-x\right)}\)

\(\Leftrightarrow\frac{x-3}{x-2}>0\)

\(\Rightarrow\orbr{\begin{cases}x-3\ge0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x\ge3\\x>2\end{cases}\Leftrightarrow}x\ge3}\)

\(\Rightarrow\orbr{\begin{cases}x-3< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 3\\x< 2\end{cases}\Leftrightarrow}x< 2}\)

Vậy ...

Cho biểu thức : A= ( 3/2x+4 + x/2-x + 2x^2+3/x^2-4 ) : (2x-1/4x-8)

a.Rút gọn A

b.Tìm giá trị của A biết |x - 1| = 3

c.Tìm x để A < 2

d.Tìm x để A = |1|

a ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(P=\dfrac{x^2-4x-4}{4-x^2}+\dfrac{3x+9}{x+2}\)

\(=\dfrac{-x^2+4x+4}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x+9}{\left(x+2\right)}\)

\(=\dfrac{-x^2+4x+4+\left(3x+9\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-x^2+4x+4+3x^2-6x+9x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x^2+7x-14}{\left(x-2\right)\left(x+2\right)}\)

b: khi x=8 thì \(P=\dfrac{2\cdot8^2+7\cdot8-14}{\left(8-2\right)\left(8+2\right)}=\dfrac{2\cdot64+56-14}{64-4}=\dfrac{17}{6}\)

1,

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

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

\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)

2.

\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

3.

Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)

4.

\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)

\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)

5.

\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)

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

\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)

\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)

a) Ta có: \(A=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\dfrac{2x-1}{4x-8}\)

\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right):\dfrac{2x-1}{4x-8}\)

\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)

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

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

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