Điều kiện xác định: \(x\ge0;x\ne9\)

1/ \(P=\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-3}{3-\sqrt{x}}-\dfrac{3\left(3\sqrt{x}-5\right)}{x-2\sqrt{x}-3}\)

\(=\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}+\dfrac{2\sqrt{x}-3}{\sqrt{x}-3}-\dfrac{9\sqrt{x}-15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x-7\sqrt{x}-6+2x-\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(5\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{5\sqrt{x}-2}{\sqrt{x}+1}\)

b) Khi \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Ta có \(P=\dfrac{5\left(\sqrt{3}+1\right)-2}{\sqrt{3}+1+1}=\dfrac{5\sqrt{3}+3}{\sqrt{3}+2}\)

c) \(P=\dfrac{5\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-7}{\sqrt{x}+1}=5-\dfrac{7}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow P\ge5-\dfrac{7}{1}=-2\)

Dấu = xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy \(P_{min}=-2\) đạt được khi \(x=0\)

Lời giải:

$\frac{\sqrt{x}+1}{\sqrt{x}+4}=\frac{\sqrt{x}+4-3}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}$

Vì $\sqrt{x}\geq 0$ nên $\sqrt{x}+4\geq 4$
$\Rightarrow \frac{3}{\sqrt{x}+4}\leq \frac{3}{4}$

$\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}\geq 1-\frac{3}{4}=\frac{1}{4}$

Vậy $M=\frac{1}{4}$

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$N=\frac{\sqrt{x}+5}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}$

Do $\sqrt{x}\geq 0$ nên $\sqrt{x}+2\geq 2$

$\Rightarrow \frac{3}{\sqrt{x}+2}\leq \frac{3}{2}$

$\Rightarrow \frac{\sqrt{x}+5}{\sqrt{x}+2}\leq 1+\frac{3}{2}=\frac{5}{2}$

Vậy $N=\frac{5}{2}$

$\Rightarrow 2M+N =2.\frac{1}{4}+\frac{5}{2}=3$

Đáp án C.

Đk: \(x\ge0\)

\(P=\dfrac{\sqrt{x}}{x+3\sqrt{x}+4}\)

\(\Leftrightarrow x.P+\sqrt{x}\left(3P-1\right)+4P=0\) (1)

Xét P=0 <=> x=0(tm)

Xét \(P\ne0\) .Coi pt (1) là phương trình ẩn \(\sqrt{x}\)

Phương trình (1) có nghiệm không âm khi \(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\S\ge0\\P\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-7P^2-6P+1\ge0\\\dfrac{1-3P}{P}\ge0\\4\ge0\left(lđ\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-1\le P\le\dfrac{1}{7}\\0< P\le\dfrac{1}{3}\end{matrix}\right.\) \(\Rightarrow0< P\le\dfrac{1}{7}\)

Kết hợp với P=0 \(\Rightarrow0\le P\le\dfrac{1}{7}\)

Có \(\dfrac{1}{7}>0\) => maxP=\(\dfrac{1}{7}\). Thay \(P=\dfrac{1}{7}\) vào (1) tìm được x=4 (tm)

minP=0 <=> x=0

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{2}{\sqrt{x}-2}-\dfrac{4\sqrt{x}}{x-4}\)

\(=\dfrac{x-2\sqrt{x}+2\sqrt{x}+4-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)

Với \(x< 9\) biểu thức này chỉ có max, ko có min

Để có min thì cần \(x>9\)

\(P=\dfrac{x-5}{\sqrt{x}-3}=\dfrac{x-9+4}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+4}{\sqrt{x}-3}\)

\(P=\sqrt{x}+3+\dfrac{4}{\sqrt{x}-3}=\sqrt{x}-3+\dfrac{4}{\sqrt{x}-3}+6\)

\(P\ge2\sqrt{\left(\sqrt{x}-3\right).\dfrac{4}{\sqrt{x}-3}}+6=10\)

\(P_{min}=10\) khi \(\sqrt{x}-3=\dfrac{4}{\sqrt{x}-3}\Leftrightarrow\sqrt{x}-3=2\Rightarrow x=25\)

Nếu chưa học BĐT Cô-si như cách làm trên thì:

\(P=\dfrac{x-5}{\sqrt{x}-3}=\dfrac{x-10\sqrt{x}+25+10\sqrt{x}-30}{\sqrt{x}-3}\)

\(P=\dfrac{\left(\sqrt{x}-5\right)^2+10\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-5\right)^2}{\sqrt{x}-3}+10\)

Do \(x>9\Rightarrow\sqrt{x}>3\Rightarrow\sqrt{x}-3>0\Rightarrow\dfrac{\left(\sqrt{x}-5\right)^2}{\sqrt{x}-3}>0\)

\(\Rightarrow P\ge10\)

a: TXĐ: D=[0;+\(\infty\))\{1}

Ta có: \(P=\left(\dfrac{3}{\sqrt{x}+1}-\dfrac{1}{x-1}\right):\dfrac{1}{\sqrt{x}+1}\)

\(=\dfrac{3\sqrt{x}-3-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)

\(=\dfrac{3\sqrt{x}-4}{\sqrt{x}-1}\)

Ta có: \(P=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=1-\dfrac{8}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\)

\(\Rightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\)

\(\Leftrightarrow-\dfrac{8}{\sqrt{x}+3}\ge-\dfrac{8}{3}\)

\(\Leftrightarrow1-\dfrac{8}{\sqrt{x}+3}\ge1-\dfrac{8}{3}=-\dfrac{5}{3}\)

\(\Leftrightarrow P\ge-\dfrac{5}{3}\)

Dấu "=" xảy ra khi x = 0

Vậy \(P_{min}=-\dfrac{5}{3}\) khi x = 0

Điều kiện: \(x\ge0\).

Ta biến đổi: \(P=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}=1-\dfrac{8}{\sqrt{x}+3}\).

Ta có: \(\sqrt{x}+3\ge3\Rightarrow-\dfrac{8}{\sqrt{x}+3}\ge-\dfrac{8}{3}\)

\(\Leftrightarrow P=1-\dfrac{8}{\sqrt{x}+3}\ge1-\dfrac{8}{3}=-\dfrac{5}{3}\)

Vậy: Giá trị nhỏ nhất của \(P\) là \(-\dfrac{5}{3}\). Dấu đẳng thức xảy ra khi và chỉ khi \(x=0\)