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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 : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a) P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\) với x > 0 và x≠4
=\(\left(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{x-4}\right).\dfrac{x-4}{2\sqrt{x}}\)
=\(\left(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\right)\).\(\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{2x}{x-4}.\dfrac{x-4}{2\sqrt{x}}\)
=\(\dfrac{x}{\sqrt{x}}\)
b) \(\dfrac{x}{\sqrt{x}}\) >3
<=> x> \(3\sqrt{x}\)
<=> x>9
a: \(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b: Để P>3 thì \(\sqrt{x}>3\)
hay x>9
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
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/ \(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{6}=\dfrac{-1}{\sqrt{x}-2}\)
b/ \(A>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0\)
Ta thấy: - 1 < 0 nên để A > 0 thì:
\(\sqrt{x}-2< 0\)\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
kết hợp với đkxđ: => \(0\le x< 4\)
Bài 1:
\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)
Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Bài 2:
\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
a: \(B=\dfrac{x+4\sqrt{x}+4-\left(x-4\sqrt{x}+4\right)+4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(4-x\right)}{\left(3-3\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\dfrac{5x+4\sqrt{x}+4-x+4\sqrt{x}-4}{1}\cdot\dfrac{-\sqrt{x}}{\left(3-3\sqrt{x}\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x+8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{3\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{3\sqrt{x}-3}\)
b: Để B>0 thì \(3\sqrt{x}-3>0\)
hay x>1
c: Để B là số nguyên thì \(4\sqrt{x}⋮3\sqrt{x}-3\)
\(\Leftrightarrow12\sqrt{x}-12+12⋮3\sqrt{x}-3\)
\(\Leftrightarrow3\sqrt{x}-3\in\left\{1;-1;2;-2;3;-3;4;6;12\right\}\)
hay \(x\in\left\{\dfrac{16}{9};\dfrac{4}{9};\dfrac{25}{9};\dfrac{1}{9};\dfrac{49}{9};9;25\right\}\)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
b) \(P=\sqrt{x}>3\Leftrightarrow x>9\)
a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{2\sqrt{x}}\)
\(=\sqrt{x}\)
b: Để P>3 thì x>9