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Với x >= 0 ; x khác 9
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\frac{-3\sqrt{x}-3}{x-9}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}\)
\(\frac{B}{A}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{-3}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Kết hợp đk vậy 0 =< x < 9
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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\(1,A=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
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 \(\left(\sqrt{x}-1\right)^2=3\Rightarrow\sqrt{x}=\pm\sqrt{3}+1\)
\(\Rightarrow x=\left(\pm\sqrt{3}+1\right)^2=4\pm2\sqrt{3}\)
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\(=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)}\right):\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\right)\)
\(=\frac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(x=4\Rightarrow B=\frac{4+2+1}{2}=\frac{7}{2}\)
\(B=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{x}}}+1=3\)
\(B_{min}=3\) khi \(x=1\)
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\(Z=\frac{\sqrt{x}-5}{\sqrt{x}+2}=1-\frac{7}{\sqrt{x}+2}\ge1-\frac{7}{2}=-\frac{5}{2}\)
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Lời giải:
Có:
$\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}$
Ta thấy:
$\sqrt{x}\geq 0, \forall x\geq 0; x\neq 4\Rightarrow \sqrt{x}+2\geq 2$
$\Rightarrow \frac{1}{\sqrt{x}+2}\leq \frac{1}{2}$
$\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}+2}=1-\frac{1}{\sqrt{x}+2}\geq 1-\frac{1}{2}=\frac{1}{2}$
Vậy GTNN của biểu thức là $\frac{1}{2}$ xảy ra khi $\sqrt{x}=0$ hay $x=0$
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(P=A\cdot B\)
\(=\dfrac{\sqrt{x}+7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}+7}{x+3\sqrt{x}+2}\)
Đề thiếu rồi bạn
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\(B=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}+1}\)
\(B=\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\)
\(B=\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}+1\right)-\left(x-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x^2+x\sqrt{x}+\sqrt{x}+1-x^2+2x-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x\sqrt{x}+\sqrt{x}+2x}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(x+1+2\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
tiếp tục của bạn @Bastkoo nhé
\(B=\frac{\sqrt{x}\left(x+2\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(< =>B=\frac{\sqrt{x}}{\sqrt{x}-1}\)