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a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )
a)A \(=\dfrac{\sqrt{x}+1}{x+4\sqrt{x}+4}:\left(\dfrac{x}{x+2\sqrt{x}}+\dfrac{x}{\sqrt{x}+2}\right)\)
A=\(\dfrac{\sqrt{x}+1}{\sqrt{x^2}+2.2.\sqrt{x}+2^2}:\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{x}{\sqrt{x}+2}\right)\)
A\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}:\left(\dfrac{x+x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\)
A\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}.\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x+x\sqrt{x}}\)
A\(=\dfrac{\left(\sqrt{x}+1\right)\left[\sqrt{x}\left(\sqrt{x}+2\right)\right]}{\left(\sqrt{x}+2\right)^2.\left(x+x\sqrt{x}\right)}\)
A\(=\dfrac{\left(\sqrt{x}+1\right).\sqrt{x}}{\left(\sqrt{x}+2\right).\left[x\left(\sqrt{x}+1\right)\right]}\)
A\(=\dfrac{\sqrt{x}}{\left(\sqrt{x}+2\right).x}\)
A\(=\dfrac{1}{\left(\sqrt{x}+2\right)\sqrt{x}}\)
A\(=\dfrac{1}{x+2\sqrt{x}}\)
b) \(\dfrac{1}{x+2\sqrt{x}}\ge\dfrac{1}{3\sqrt{x}}\)
\(\Leftrightarrow\dfrac{1}{x+2\sqrt{x}}-\dfrac{1}{3\sqrt{x}}\ge0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-x-2\sqrt{x}}{\left(x+2\sqrt{x}\right)\left(3\sqrt{x}\right)}\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-x}{3x\sqrt{x}+6x}\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(3x+6\sqrt{x}\right)}\ge0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{3x+6\sqrt{x}}\ge0\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
A=(\(\dfrac{\left(x+4\right)}{3\left(x+2\right)}-\dfrac{1}{\left(x+2\right)^2}\))(\(\dfrac{x+5+x-1}{x+5}\))
A=\(\dfrac{\left(x+4\right)\left(x+2\right)-3}{3\left(x+2\right)^2}\cdot\dfrac{2x+2}{x+5}\)
A=\(\dfrac{x^2+2x+4x+8-3}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+5}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+9-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{\left(x+3\right)^2-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{2\left(x+3-2\right)\left(x+3+2\right)}{3\left(x-2\right)\left(x+5\right)}\)
A=\(\dfrac{2\left(x+1\right)}{3\left(x-2\right)}\)
a: \(P=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b: ĐểP<15/4 thì P-15/4<0
\(\Leftrightarrow4\left(3\sqrt{x}+8\right)-15\left(\sqrt{x}+2\right)< 0\)
=>12 căn +32-15 căn x+30<0
=>-3 căn x<-62
=>căn x>62/3
=>x>3844/9
a)đkxđ:\(x\ge0,x\ne4\)
P=\(\left(\dfrac{2}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{\sqrt{x+2}}\)
\(=\left(\dfrac{2}{\left(\sqrt{x-2}\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\right):\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\sqrt{x}+2\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
a) \(P=\dfrac{x\left(x-4\right)+4}{x^2-4}\)
\(P=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(P=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(P=\dfrac{x-2}{x+2}\)
b) \(P=\dfrac{x-2}{x+2}\)=\(\dfrac{1001}{1003}\)
\(=>1003x-2006=1001x+2002\)
\(2x=4008=>x=2004\)
a) ĐK: \(x\ne2;x\ne-2\)
\(P=\dfrac{x\left(x-4\right)+4}{x^2-4} =\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
b) \(P=\dfrac{1001}{1003}\Leftrightarrow\dfrac{x-2}{x+2}=\dfrac{1001}{1003}\Rightarrow1003x-2006=1001x+2002\)
\(\Leftrightarrow2x=4008\Leftrightarrow x=2004\) (tmđk)