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

\(=\dfrac{1}{x\left|x-1\right|}\)

\(\dfrac{2\sqrt{x}}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\left(ĐK:x\ge0;x\ne4;x\ne9\right)\)

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

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

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

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

\(=\dfrac{x-\sqrt{x}+7}{x-5\sqrt{x}+6}\)

#Urushi

\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\) (ĐK: \(x\ne9;x\ne4;x\ge0\))

\(=\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

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

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

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

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

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

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

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

ĐKXĐ: x>0; x<>4; x<>1

\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{4\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{4}\)

\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\right)\left(ĐKXĐ:x>0;x\ne1;x\ne4\right)\)

\(=\left[\dfrac{\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-3}{x-2\sqrt{x}}\right]:\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)

\(=\left[\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\dfrac{x-4-\left(x-4\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{x-3\sqrt{x}-\sqrt{x}+3}{-4+4\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)}{4\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}\)

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

#Urushi☕

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

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

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

\(=\dfrac{4x\sqrt{x}-7x+3\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

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

Để A>1 thì A-1>0

\(\Leftrightarrow\dfrac{4\sqrt{x}-3-\sqrt{x}+2}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{3\sqrt{x}-1}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1\le0\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x\le\dfrac{1}{9}\\x>4\end{matrix}\right.\)

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

b.

\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

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

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

b: Ta có: \(\sqrt{4x+8}+\dfrac{1}{3}\sqrt{9x+18}=3\sqrt{\dfrac{x+2}{4}}+\sqrt{2}\)

\(\Leftrightarrow2\sqrt{x+2}+\dfrac{1}{3}\cdot3\sqrt{x+2}-\dfrac{3}{2}\sqrt{x+2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}\cdot\dfrac{3}{2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{2\sqrt{2}}{3}\)

\(\Leftrightarrow x+2=\dfrac{8}{9}\)

hay \(x=-\dfrac{10}{9}\)

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

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

a) Ta có: \(A=\dfrac{2}{x-1}\cdot\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)

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

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

b) Ta có: \(\left(x^2-4\right)\cdot\sqrt{\dfrac{9}{x^2-4x+4}}\)

\(=\dfrac{\left(x-2\right)\left(x+2\right)\cdot3}{\left(x-2\right)^2}\)

\(=\dfrac{3x+6}{x-2}\)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)