Đề sai rồi bạn

\(A=\sqrt{x}+\dfrac{2}{\sqrt{x}}\ge2\cdot\sqrt{\sqrt{x}\cdot\dfrac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu '=' xảy ra khi \(\sqrt{x}\cdot\sqrt{x}=2\)

hay \(x=2\)

a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)

\(\Rightarrow A=\frac{2+1}{2+2}=\frac{3}{4}\)

Vậy với x = 4 thì A = 3/4 

b, \(B=\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}+5}{x-1}=\frac{3\left(\sqrt{x}+1\right)-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{\sqrt{x}+1}\)( đpcm )

với x=4(t/m DK)

=>\(\sqrt{x}\)=2

thay\(\sqrt{x}\)=2 vào biểu thức A ta được

A=(2+1)/(2+2)

A=3/4

\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)

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

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)

+ Với \(x+\sqrt{x}+1=1\)

\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)

+ Với \(x+\sqrt{x}+1=2\)

\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)

Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)

 P = A.B = \(\dfrac{x-7}{\sqrt{x}+2}=\dfrac{\left(x-4\right)-3}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)-3}{\sqrt{x}+2}\)

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

\(P\inℤ\) <=> x là số chính phương và \(\dfrac{3}{\sqrt{x}+2}\inℤ\)

mà \(\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow x=1\) (thỏa)

Vậy x = 1 thì P \(\inℤ\)

Lời giải:

$M=\frac{2(\sqrt{x}-3)+7}{\sqrt{x}-3}=2+\frac{7}{\sqrt{x}-3}$

Để $M$ nguyên thì $\frac{7}{\sqrt{x}-3}$

Với $x$ nguyên không âm thì điều này xảy ra khi mà $\sqrt{x}-3$ là ước của $7$

$\Rightarrow \sqrt{x}-3\in\left\{\pm 1; \pm 7\right\}$

$\Rightarrow \sqrt{x}\in \left\{4; 2; 10; -4\right\}$

Vì $\sqrt{x}\geq 0$ nên $\sqrt{x}\in \left\{4; 2; 10\right\}$

$\Rightarrow x\in \left\{16; 4; 100\right\}$ (tm)

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)

\(a,ĐK:x\ge1;x\ne3\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\)

`B=(x+sqrtx+5)/(sqrtx+1)=(sqrtx(sqrtx+1)+4)/(sqrtx+1)=sqrtx+4/(sqrtx+1)=[(sqrtx+1)+4/(sqrtx+1)]-1>=2\sqrt((sqrtx+1). 4/(sqrtx+1))-1=3`

Dấu "=" xảy ra `<=>x=1`

Vậy `B_(min)=3<=>x=1`