Đề sai rồi bạn

Lời giải:

ĐK: $x\geq 0; x\neq 4; x\neq 9$

a) 

\(P=\frac{2\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{(2\sqrt{x}+1)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}-\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-2)}\)

\(=\frac{2\sqrt{x}-9+(2\sqrt{x}+1)(\sqrt{x}-2)-(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{x-\sqrt{x}-2}{(\sqrt{x}-3)(\sqrt{x}-2)}\)

\(=\frac{(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Với $x$ nguyên, để $P$ nguyên thì $\sqrt{x}-3$ phải là ước nguyên của $4$

Mà $\sqrt{x}-3\geq -3$ nên:

$\Rightarrow \sqrt{x}-3\in\left\{\pm 1;\pm 2;4\right\}$

$\Rightarrow x\in \left\{4;16;1;25;49\right\}$ (đều thỏa mãn.

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

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

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)

\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)

Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)

\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)

\(a,\) Rút gọn 

\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)

\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)

\(=\dfrac{3}{\sqrt{7}-2}+3-\sqrt{7}\)

\(=\dfrac{3+\left(3-\sqrt{7}\right)\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)

\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)

\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)

\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)

\(=5\)

Vậy \(A=5\)

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

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

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

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

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

Vậy \(B=\sqrt{x}-1\)

\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)

\(\Leftrightarrow\sqrt{x}< 6\)

\(\Leftrightarrow x< 36\)

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

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

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

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

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

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

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

b)Để P<4 \(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}< 4\) \(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-4< 0\) \(\Leftrightarrow\dfrac{\sqrt{x}+2-4\left(\sqrt{x}-1\right)}{\sqrt{x}-1}< 0\)

\(\Leftrightarrow\dfrac{6-3\sqrt{x}}{\sqrt{x}-1}< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-3\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}6-3\sqrt{x}< 0\\\sqrt{x}-1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}< 1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}< 1\\\sqrt{x}>2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0\le x< 1\\x>4\end{matrix}\right.\)

Vậy...

c)\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\) \(=1+\dfrac{3}{\sqrt{x}-1}\)

Để P nguyên khi \(\dfrac{3}{\sqrt{x}-1}\) nguyên

\(x\in Z\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}\in Z\\\sqrt{x}\in I\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-1\in Z\\\sqrt{x}-1\in I\end{matrix}\right.\)

Tại \(\sqrt{x}-1\in I\Rightarrow\dfrac{3}{\sqrt{x}-1}\notin Z\) (L)

Tại\(\sqrt{x}-1\in Z\) .Để \(\dfrac{3}{\sqrt{x}-1}\in Z\)

\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;-2;4\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;4\right\}\) \(\Leftrightarrow x\in\left\{0;4;16\right\}\) (tm)

câu c là sao vậy ạ??

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

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

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

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

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

\(A\ge\dfrac{1}{3\sqrt{x}}\Leftrightarrow\dfrac{1}{\sqrt{x}\left(\sqrt{x}+2\right)}\ge\dfrac{1}{3\sqrt{x}}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}\ge\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2\le3\)

\(\Rightarrow x\le1\)

Kết hợp ĐKXĐ \(\Rightarrow0< x\le1\)

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

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

=\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\) 

b)Ta có A = \(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\)=2+\(\dfrac{2}{\sqrt{x}-1}\)

Để A nguyên thì \(\sqrt{x}-1\)∈Ư₍₂₎

⇒x∈{4;0;9}