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ĐKXĐ của cả A và B : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(B=\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\frac{\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x+4\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{x-\sqrt{x}+5\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-5}\)
\(M=\frac{B}{A}=\frac{\frac{\sqrt{x}-1}{\sqrt{x}-5}}{\frac{\sqrt{x}+2}{\sqrt{x}-5}}=\frac{\sqrt{x}-1}{\sqrt{x}-5}\times\frac{\sqrt{x}-5}{\sqrt{x}+2}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
ĐKXĐ của M : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)
\(M\times\left(\sqrt{x}+2\right)\ge3x-3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}\times\left(\sqrt{x}+2\right)\ge3x-3\)( ĐK : \(\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\))
\(\Leftrightarrow\sqrt{x}-1\ge3x-3\)
\(\Leftrightarrow3x-\sqrt{x}-3+1\ge0\)
\(\Leftrightarrow3x-\sqrt{x}-2\ge0\)
\(\Leftrightarrow3x-3\sqrt{x}+2\sqrt{x}-2\ge0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}+2\right)\ge0\)
Dễ dàng nhận thấy \(3\sqrt{x}+2\ge2>0\forall x\ge0\)
\(\Rightarrow\sqrt{x}-1\ge0\)
\(\Leftrightarrow x\ge1\)
Kết hợp với điều kiện => Với 0 ≤ x ≤ 1 thì thỏa mãn đề bài
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
a) Thay x = 25 vào biểu thức A , ta có
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
b) \(B=\frac{x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{x+1+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Ta có : \(x=25\Rightarrow\sqrt{x}=5\)
Thay vào biểu thức A ta được :
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
Vậy với x = 25 thì A = 3/4
b, Với \(x\ge0;x\ne1\)
\(B=\frac{x-5}{x-1}-\frac{2}{\sqrt{x}+1}+\frac{4}{\sqrt{x}-1}\)
\(=\frac{x-5-2\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)}{x-1}=\frac{x-5-2\sqrt{x}+2+4\sqrt{x}+4}{x-1}\)
\(=\frac{x+1+2\sqrt{x}}{x-1}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}\pm1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c, Ta có P = A/B hay \(P=\frac{\sqrt{x}-2}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(\sqrt{P}< \frac{1}{2}\)hay \(\sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+1}}< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}-\frac{1}{4}< 0\Leftrightarrow\frac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow3\sqrt{x}-9>0\)do \(4\left(\sqrt{x}+1\right)>0\)
\(\Leftrightarrow3\sqrt{x}>9\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
\(1,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
2, Với x>1 ta có \(\frac{1}{A}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\frac{3}{\sqrt{x}-1}+3\)
Áp dụng bđt AM-GM ta có
\(\frac{1}{A}\ge2\sqrt{\left(\sqrt{x}-1\right).\frac{3}{\sqrt{x}-1}}+3=2\sqrt{3}+3\)
Dấu "=" xảy ra khi \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)
\(Z=\frac{\sqrt{x}-5}{\sqrt{x}+2}=1-\frac{7}{\sqrt{x}+2}\ge1-\frac{7}{2}=-\frac{5}{2}\)