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.

lam gjup vs mn oi

1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)

\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)

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

\(=\frac{-2}{\sqrt{x}-2}.\frac{1}{6}=-\frac{1}{3\left(\sqrt{x}-2\right)}\)

2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)

                             \(\Rightarrow x>\frac{121}{36}\)

sai r bn oi

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

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

b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:

\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)

a. ĐKXĐ \(x\ge0\)và \(x\ne9\)

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

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)

\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)

Vậy với \(0\le x< \frac{9}{4}\)thì K<-1

c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)

Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)

\(\Rightarrow K\ge-3\)

Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)