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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
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
a)\(ĐKXĐ\Leftrightarrow\begin{cases}\sqrt{x}\ge0\\\sqrt{x}-1\ne0\end{cases}\Leftrightarrow\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(A=\frac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)+1\cdot\left(\sqrt{x}-1\right)-3\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
b)\(S=A\cdot B\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+2+1}{\sqrt{x}+2}\)
\(=1+\frac{1}{\sqrt{x}+2}\)
Để S đạt GTLN thì \(\frac{1}{\sqrt{x}+2}\) đạt GTLN
\(\frac{1}{\sqrt{x}+2}\) đạt GTLN \(\Leftrightarrow\sqrt{x}+2\) đạt GTNN
GTNN \(\sqrt{x}+2\) là 2 \(\Leftrightarrow x=0\)
Vậy GTLN của S là \(\frac{3}{2}\Leftrightarrow x=0\)
a/ \(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-\frac{3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) \(\left(ĐK:x\ge0;x\ne1\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+\sqrt{x}-1-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
a) x = 16 (tm) => A = \(\frac{\sqrt{16}-2}{\sqrt{16}+1}=\frac{4-2}{4+1}=\frac{2}{5}\)
b) B = \(\left(\frac{1}{\sqrt{x}+5}-\frac{x+2\sqrt{x}-5}{25-x}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
B = \(\frac{\sqrt{x}-5+x+2\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
B = \(\frac{x+3\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{x+5\sqrt{x}-2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
B = \(\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
c) P = \(\frac{B}{A}=\frac{\sqrt{x}-2}{\sqrt{x}+2}:\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
=> \(P\left(\sqrt{x}+2\right)\ge x+6\sqrt{x}-13\)
<=> \(\frac{\sqrt{x}+1}{\sqrt{x}+2}.\left(\sqrt{x}+2\right)-x-6\sqrt{x}+13\ge0\)
<=> \(-x-6\sqrt{x}+13+\sqrt{x}+1\ge0\)
<=> \(-x-5\sqrt{x}+14\ge0\)
<=> \(x+5\sqrt{x}-14\le0\)
<=> \(x+7\sqrt{x}-2\sqrt{x}-14\le0\)
<=> \(\left(\sqrt{x}+7\right)\left(\sqrt{x}-2\right)\le0\)
Do \(\sqrt{x}+7>0\) với mọi x => \(\sqrt{x}-2\le0\)
<=> \(\sqrt{x}\le2\) <=> \(x\le4\)
Kết hợp với Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)25
và x thuộc Z => x = {0; 1; 2; 3}
d) M = \(3P\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\) <=>M = \(3\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\)
M = \(\frac{3\sqrt{x}+3}{x+\sqrt{x}+4}=\frac{x+\sqrt{x}+4-x+2\sqrt{x}-1}{\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{15}{4}}=1-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}}\le1\)(Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\) <=> \(x=1\)
Vậy MaxM = 1 khi x = 1
ĐKXĐ \(x\ge0;x\ne4\)
1. Với x = 25 :
\(A=\dfrac{\sqrt{25}+1}{25-4}=\dfrac{2}{7}\)
2. \(B=\dfrac{18-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}+\dfrac{4\left(\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(\sqrt{x+2}\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{18-\sqrt{x}-4\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
3.\(P=A.B=\dfrac{\sqrt{x}+1}{x-4}.\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\)
<=> 4P = \(\dfrac{4\sqrt{x}+4}{\left(\sqrt{x}+2\right)^2}=\dfrac{x+4\sqrt{x}+4-x}{\left(\sqrt{x}+2\right)^2}=1-\dfrac{x}{\left(\sqrt{x}+2\right)^2}\le1\)(Do \(x\ge0\))
<=> \(P\le\dfrac{1}{4}\)("Dấu "=" xảy ra <=> x = 0)