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đk x >= 0 ; x khác 1/4
Ta có \(^{P=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}+1}}=\dfrac{5\sqrt{x}+1}{2\sqrt{x}+1}\)
\(\Rightarrow5\sqrt{x}+1⋮2\sqrt{x}+1\Leftrightarrow10\sqrt{x}+2⋮2\sqrt{x}+1\)
\(\Leftrightarrow5\left(2\sqrt{x}+1\right)-3⋮2\sqrt{x}+1\Rightarrow2\sqrt{x}+1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| \(2\sqrt{x}+1\) | 1 | -1 | 3 | -3 |
| x | 0 | loại | 1 | loại |
Lời giải:
a. ĐKXĐ: $x>1$
\(B=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{(\sqrt{x+1}+\sqrt{x-1})^2}{2}=x+\sqrt{x^2-1}\)
b.
\(B=\frac{a^2+b^2}{2ab}+\sqrt{\frac{a^2+2ab+b^2}{2ab}.\frac{a^2-2ab+b^2}{2ab}}\)
\(=\frac{a^2+b^2}{2ab}+\sqrt{\frac{(a+b)^2(a-b)^2}{(2ab)^2}}=\frac{a^2+b^2}{2ab}+\frac{|a-b||a+b|}{|2ab|}=\frac{a^2+b^2}{2ab}+\frac{a^2-b^2}{2ab}=\frac{a}{b}\)
c.
$B\leq 1\Leftrightarrow (x-1)+\sqrt{x^2-1}\leq 0$
$\Leftrightarrow \sqrt{x-1}(\sqrt{x-1}+\sqrt{x+1})\leq 0$
$\Leftrightarrow \sqrt{x-1}\leq 0$
Mà $\sqrt{x-1}>0$ với mọi $x<1$ nên điều này vô lý)
Vậy không tồn tại $x$ thỏa đkđb
d.
$B=2\Leftrightarrow x+\sqrt{x^2-1}=2$
$\Leftrightarrow \sqrt{x^2-1}=2-x$
\(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ x^2-1=(2-x)^2=x^2-4x+4\end{matrix}\right.\)
\(\Rightarrow x=\frac{5}{4}\)
Thử lại thấy thỏa mãn
Vậy......
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
\(A=\left(\dfrac{1}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{1+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\inƯ\left(2\right)\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{3;1;4;0\right\}\)
=>\(x\in\left\{9;1;16;0\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{9;16\right\}\)
c: A<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
Kết hợp ĐKXĐ, ta được: 0<x<4 và x<>1
a: \(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
Khi x=16 thì \(B=\dfrac{2\cdot4+2}{\left(4-2\right)\left(4+2\right)}=\dfrac{10}{2\cdot6}=\dfrac{10}{12}=\dfrac{5}{6}\)
b: P=B/A
\(=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{2}{\sqrt{x}+2}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
c: P<1
=>P-1<0
=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên
nên \(x\in\left\{0;1;2;3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;1;2;3\right\}\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
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 \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)
`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`
`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`
`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`
`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`
`B=2/(3-sqrtx)`
`B>1/2`
`<=>2/(3-sqrtx)-1/2>0`
`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`
`<=>(sqrtx+1)/(2(3-sqrtx))>0`
Mà `sqrtx+1>=1>0`
`<=>2(3-sqrtx)>0`
`<=>3-sqrtx>0`
`<=>sqrtx<3`
`<=>x<9`
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{x-4}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) Với \(x\ge0;x\ne1;x\ne4\):
Thay \(x=3+2\sqrt{2}\) vào \(P\), ta được:
\(P=\dfrac{\sqrt{3+2\sqrt{2}}+2}{\sqrt{3+2\sqrt{2}}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}+2}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(=\dfrac{\sqrt{2}+1+2}{\sqrt{2}+1-1}\)
\(=\dfrac{\sqrt{2}+3}{\sqrt{2}}\)
\(=\dfrac{2+3\sqrt{2}}{2}\)
c) Với \(x\ge0;x\ne1;x\ne4\),
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên
\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)
\(\Rightarrow x\in\left\{4;16;0\right\}\)
Kết hợp với ĐKXĐ của \(x\), ta được:
\(x\in\left\{0;16\right\}\)
Vậy: ...
\(\text{#}Toru\)
Lời giải:
$5A+B=\frac{5\sqrt{x}+1}{2\sqrt{x}+1}$
$2(5A+B)=\frac{10\sqrt{x}+2}{2\sqrt{x}+1}=\frac{5(2\sqrt{x}+1)-3}{2\sqrt{x}+1}=5-\frac{3}{2\sqrt{x}+1}$
$5A+B$ nguyên
$\Rightarrow 2(5A+B)$ nguyên
$\Leftrightarrow 5-\frac{3}{2\sqrt{x}+1}$ nguyên
$\Leftrightarrow \frac{3}{2\sqrt{x}+1}$ nguyên
Ta thấy: $\frac{3}{2\sqrt{x}+1}\leq 3$ với mọi $x\geq 0$ và $\frac{3}{2\sqrt{x}+1}>0$ với mọi $x\geq 0$
Do đó $\frac{3}{2\sqrt{x}+1}$ nguyên thì nhận các giá trị $1,2,3$
$\Leftrightarrow x=0; \frac{1}{16}; 1$