a.

Đặt \(\sqrt{x}+1=t\Rightarrow t\ge3\)

\(\sqrt{x}=t-1\)

\(\Rightarrow D=\dfrac{\left(t-1\right)^2-\left(t-1\right)+2}{t}=\dfrac{t^2-3t+4}{t}=t+\dfrac{4}{t}-3\)

\(D=\dfrac{4t}{9}+\dfrac{4}{t}+\dfrac{5t}{9}-3\ge2\sqrt{\dfrac{16t}{9t}}+\dfrac{5}{9}.3-3=\dfrac{4}{3}\)

\(D_{min}=\dfrac{4}{3}\) khi \(t=3\) hay \(x=4\)

b.

Đặt \(\sqrt{x}+2=t\Rightarrow t\ge4\)

\(\Rightarrow\sqrt{x}=t-2\)

\(M=\dfrac{\left(t-2\right)^2+8}{t}=\dfrac{t^2-4t+12}{t}=t+\dfrac{12}{t}-4\)

\(M=\dfrac{3t}{4}+\dfrac{12}{t}+\dfrac{1}{4}t-4\)

\(M\ge2\sqrt{\dfrac{36t}{4t}}+\dfrac{1}{4}.4-4=3\)

\(M_{min}=3\) khi \(t=4\) hay \(x=4\)

b.

\(M=\frac{x+8}{\sqrt{x}+2}=\frac{\sqrt{x}(\sqrt{x}+2)-2(\sqrt{x}+2)+12}{\sqrt{x}+2}=\sqrt{x}-2+\frac{12}{\sqrt{x}+2}\)

\(=\frac{3}{4}(\sqrt{x}+2)+\frac{12}{\sqrt{x}+2}+\frac{\sqrt{x}}{4}-\frac{7}{2}\)

\(\geq 2\sqrt{\frac{3}{4}.12}+\frac{\sqrt{4}}{4}-\frac{7}{2}=3\) (theo AM-GM)

Vậy $M_{\min}=3$ khi $x=4$

Có bài ngược của bài này, bạn đăng và đã có lời giải thì chỉ cần đảo lại đáp án là được.

\(E=\sqrt{x}+\dfrac{4}{\sqrt{x}}-2=\dfrac{4\sqrt{x}}{9}+\dfrac{4}{\sqrt{x}}+\dfrac{5}{9}.\sqrt{x}-2\)

\(E\ge2\sqrt{\dfrac{16\sqrt{x}}{9\sqrt{x}}}+\dfrac{5}{9}.\sqrt{9}-2=\dfrac{7}{3}\)

\(E_{min}=\dfrac{7}{3}\) khi \(x=9\)

\(F=3\sqrt{x}+\dfrac{1}{\sqrt{x}}+1=2\sqrt{x}+\dfrac{1}{\sqrt{x}}+\sqrt{x}+1\)

\(F\ge2\sqrt{\dfrac{2\sqrt{x}}{\sqrt{x}}}+1.\sqrt{\dfrac{1}{2}}+1=\dfrac{2+5\sqrt{2}}{2}\)

\(F_{min}=\dfrac{2+5\sqrt{2}}{2}\) khi \(x=\dfrac{1}{2}\)

2.

\(\frac{1}{G}=\frac{2x-5\sqrt{x}+18}{\sqrt{x}}=2\sqrt{x}-5+\frac{18}{\sqrt{x}}\)

\(=2\sqrt{x}+\frac{18}{\sqrt{x}}-5\geq 2\sqrt{2.18}-5=7\) theo BĐT AM-GM

\(\Rightarrow G\leq \frac{1}{7}\) 

Vậy \(G_{\max}=\frac{1}{7}\Leftrightarrow x=9\)

1.

\(\frac{1}{K}=\frac{x-2\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}-2+\frac{4}{\sqrt{x}}\)

\(=\frac{4\sqrt{x}}{9}+\frac{4}{\sqrt{x}}+\frac{5\sqrt{x}}{9}-2\)

\(\geq 2\sqrt{\frac{4}{9}.4}+\frac{5\sqrt{9}}{9}-2=\frac{7}{3}\) (theo BĐT AM-GM)
\(\Rightarrow K\leq \frac{3}{7}\)

Vậy \(K_{\max}=\frac{3}{7}\Leftrightarrow x=9\)

a: Ta có: \(P=\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{\sqrt{x}-1}{\sqrt{x}-x}+\dfrac{\sqrt{x}+3}{x+5\sqrt{x}+6}\)

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

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

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

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

Lời giải:

ĐKXĐ: $x>0; x\neq 4$
\(A=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2}{\sqrt{x}+2}\)

\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)

Hiển nhiên $B>0$

Với $x>0; x\neq 4\Rightarrow 3(\sqrt{x}+2)\geq 6$

$\Rightarrow B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}<3$

Vậy $0< B< 3$. $B$ nguyên $\Leftrightarrow B\in\left\{1;2\right\}$

$\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}\in\left\{1;2\right\}$

$\Leftrightarrow x\in\left\{\frac{64}{9}; \frac{1}{9}\right\}$ (tm)

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(a,A=\left(\dfrac{x+14\sqrt{x}-5}{x-25}+\dfrac{\sqrt{x}}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\left(\dfrac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{x+14\sqrt{x}-5+x-5\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(\Rightarrow A=\dfrac{2x+10\sqrt{x}-\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}\left(\sqrt{x}+5\right)-\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+2}\)

\(a,P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right)\left(dkxd:x\ge0,x\ne1\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

\(=\dfrac{\sqrt{x}.\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

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

\(b,x=4+2\sqrt{3}\Rightarrow P=\dfrac{\left(4+2\sqrt{3}\right)-2}{\sqrt{4+2\sqrt{3}}}\)

\(=\dfrac{2\sqrt{3}+4-2}{\sqrt{\sqrt{3}^2+2\sqrt{3}+1}}\)

\(=\dfrac{2\sqrt{3}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{\left|\sqrt{3}+1\right|}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=2\)

a: \(P=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{x-1}\)

\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-2}{\sqrt{x}}\)

b: Khi x=4+2căn 3 thì \(P=\dfrac{2+2\sqrt{3}}{\sqrt{3}+1}=2\)