Đề sai rồi bạn

\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Lời giải:

ĐKXĐ: $x>0; x\neq 4$

Sửa lại đề 1 chút.
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\frac{2}{\sqrt{x}+2}\)

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

Với mọi $x>0$ thì hiển nhiên $B>0$. Mặt khác, $\sqrt{x}+2\geq 2$ nên $B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}=\frac{7}{3}$

Vậy $0< B\leq \frac{7}{3}$. $B$ đạt giá trị nguyên thì $B=1;2$

$B=1\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=1$

$\Leftrightarrow x=\frac{64}{9}$ (thỏa mãn)

$B=2\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=2$

$\Leftrightarrow x=\frac{1}{9}$ (thỏa mãn)

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: Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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

b: Để \(A\ge0\) thì \(\sqrt{x}-3>0\)

hay x>9

a) Đk \(x>0\)và \(x\ne4\)

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

=\(\frac{2\sqrt{x}}{x-4}\).\(\frac{\sqrt{x}-2}{\sqrt{x}}\)

=\(\frac{2}{\sqrt{x}+2}\)

b) Để \(\frac{2}{\sqrt{x}+2}>\frac{1}{2}\)

\(\Leftrightarrow\frac{4-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

\(\Leftrightarrow\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

Vì \(2\left(\sqrt{x}+2\right)>0\)

mà\(\frac{-\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}\)\(>0\)

nên \(-\sqrt{x}+2>0\)\(\Leftrightarrow x< 4\)

Vậy vs \(0< x< 4\)thì \(A>\frac{1}{2}\)

c) Để B=\(\frac{14}{3\left(\sqrt{x}+2\right)}\)đạt giá trị nguyên thì

\(3\left(\sqrt{x}+2\right)\inƯ\left(14\right)=\left(1;2;7\right)\)

+)\(3\left(\sqrt{x}+2\right)=1\)\(\Leftrightarrow\sqrt{x}+2=\frac{1}{3}\)\(\Rightarrow\)vô lý

+)\(3\left(\sqrt{x}+2\right)=2\)\(\Leftrightarrow\sqrt{x}+2=\frac{2}{3}\)\(\Rightarrow\)vô lý

+)\(3\left(\sqrt{x}+2\right)=7\)\(\Leftrightarrow\sqrt{x}+2=\frac{7}{3}\)\(\Leftrightarrow\sqrt{x}=\frac{1}{3}\)\(\Rightarrow x=\frac{1}{9}\)(TM đk)

Vậy vs \(x=\frac{1}{9}\)thì B có giá trị nguyên

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

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

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

hay \(x=2\)