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Bài làm :
1) Khi x=9 ; giá trị của A là :
\(A=\frac{\sqrt{9}}{\sqrt{9}+2}=\frac{3}{3+2}=\frac{3}{5}\)
2) Ta có :
\(B=...\)
\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{1.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
3) Ta có :
\(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+2}\div\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)
Xét :
\(\frac{A}{B}+1=\frac{4}{\sqrt{x+2}}>0\Rightarrow\frac{A}{B}>-1\)
=> Điều phải chứng minh
1, thay x=9(TMĐKXĐ) vào A ta đk:
A=\(\dfrac{\sqrt{9}}{\sqrt{9}-2}=3\)
vậy khi x=9 thì A =3
2,với x>0,x≠4 ta đk:
B=\(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
3,\(\dfrac{A}{B}>-1\) (x>0,x≠4)
⇒\(\dfrac{\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}>-1\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}>-1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>-1\)
⇒\(\sqrt{x}-2>-1\) (vì \(\sqrt{x}+2>0\))
⇔\(\sqrt{x}>1\)⇔x=1 (TM)
vậy x=1 thì \(\dfrac{A}{B}>-1\) với x>0 và x≠4
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
a) \(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1^3}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(P=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)\(P=\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)
\(P=\dfrac{2}{x+\sqrt{x}+1}\)
b) Mà với \(x\ge0\) và \(x\ne1\) thì
\(x+\sqrt{x}+1\ge0\) và \(2>0\) nên \(P>0\)
a: \(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2}=\dfrac{2}{x+\sqrt{x}+1}\)
b: x+căn x+1+1>=1>0
2>0
=>P>0 với mọi x thỏa mãn x>=0 và x<>1
a: \(T=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(T-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>T>3
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)
a: \(T=1:\left(\dfrac{x+2+x-1}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\left(\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}\right)\)
\(=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(T-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>T>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 )
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. \(B=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
b.Ta có:
\(B=\dfrac{2}{x+\sqrt{x}+1}\). Mà \(\left[{}\begin{matrix}2>0\\x+\sqrt{x}+1=\left[\left(\sqrt{x}\right)^2+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{3}{4}>0\end{matrix}\right.\)
Vậy B>0 \(\forall x\)