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12 tháng 8 2021
a: Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b: Thay \(x=\dfrac{1}{4}\) vào P, ta được:
\(P=\left(\dfrac{1}{2}-1\right):\left(\dfrac{1}{2}+1\right)=\dfrac{-1}{2}:\dfrac{3}{2}=-\dfrac{1}{3}\)
c: Ta có: \(P< \dfrac{1}{2}\)
\(\Leftrightarrow P-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
2S
1
T
20 tháng 11 2023
ĐKXĐ: \(x\ge0;x\ne25\)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}=\dfrac{\sqrt{x}-5+7}{\sqrt{x}-5}=1+\dfrac{7}{\sqrt{x}-5}\)
Để \(A\in\mathbb{Z}\) thì: \(\dfrac{7}{\sqrt{x}-5}\) nhận giá trị nguyên
\(\Rightarrow 7\vdots\sqrt{x}-5\)
\(\Rightarrow\sqrt{x}-5\inƯ\left(7\right)\)
\(\Rightarrow\sqrt{x}-5\in\left\{1;7;-1;-7\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{6;12;4;-2\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{4;6;12\right\}\)
\(\Rightarrow x\in\left\{16;36;144\right\}\left(tm\right)\)
Vậy \(A\in \mathbb{Z}\) khi \(x\in\left\{16;36;144\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 8 2021
ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)
c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)
\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)
\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)
12 tháng 8 2021
a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Thay x=36 vào A, ta được:
\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)
c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)
\(\Leftrightarrow4\sqrt{x}=2\)
hay \(x=\dfrac{1}{4}\)
LN
0
NB
0
AH
Akai Haruma
Giáo viên
17 tháng 2 2021
Lời giải:ĐK: $a\geq 0; a\neq 9; a\neq 4$
a)
\(A=\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{\sqrt{a}+3}{\sqrt{a}-2}+\frac{2\sqrt{a}+1}{\sqrt{a}-3}\)
\(\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{(\sqrt{a}+3)(\sqrt{a}-3)}{(\sqrt{a}-2)(\sqrt{a}-3)}+\frac{(2\sqrt{a}+1)(\ \sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}\)
\(=\frac{2\sqrt{a}-9-(a-9)+(2a-3\sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{a-\sqrt{a}-2}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)
b) Để \(A< 1\Leftrightarrow \frac{\sqrt{a}+1}{\sqrt{a}-3}<1\Leftrightarrow 1+\frac{4}{\sqrt{a}-3}<1\)
\(\Leftrightarrow \frac{4}{\sqrt{a}-3}< 0\Leftrightarrow \sqrt{a}-3< 0\Leftrightarrow 0\leq a< 9\)
Kết hợp ĐKXĐ: suy ra $0\leq a< 9; a\neq 4$
c) Với $a$ nguyên, \(A=1+\frac{4}{\sqrt{a}-3}\in\mathbb{Z}\Leftrightarrow 4\vdots \sqrt{a}-3\)
$\Rightarrow \sqrt{a}-3\in\left\{\pm 1; \pm 2;\pm 4\right\}$
$\Rightarrow a\in\left\{4;16; 1;25; 49\right\}$
Kết hợp ĐKXĐ suy ra $a\in\left\{16;1;25;49\right\}$
17 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{4;9\right\}\end{matrix}\right.\)
a) Ta có: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}+1}{3-\sqrt{a}}\)
\(=\dfrac{\left(2\sqrt{a}-9\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}-\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}+\dfrac{\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{2\sqrt{a}-9-\left(a-9\right)+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{2a-\sqrt{a}-11-a+9}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-2\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)+\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)
b) Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-1< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-\dfrac{\sqrt{a}-3}{\sqrt{a}-3}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\)
\(\Leftrightarrow\dfrac{4}{\sqrt{a}-3}< 0\)
mà 4>0
nên \(\sqrt{a}-3< 0\)
\(\Leftrightarrow\sqrt{a}< 3\)
hay a<9
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)
Vậy: Để A<1 thì \(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)
c) Để A nguyên thì \(\sqrt{a}+1⋮\sqrt{a}-3\)
\(\Leftrightarrow\sqrt{a}-3+4⋮\sqrt{a}-3\)
mà \(\sqrt{a}-3⋮\sqrt{a}-3\)
nên \(4⋮\sqrt{a}-3\)
\(\Leftrightarrow\sqrt{a}-3\inƯ\left(4\right)\)
\(\Leftrightarrow\sqrt{a}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
mà \(\sqrt{a}-3\ge-3\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-3\in\left\{1;-1;2;-2;4\right\}\)
\(\Leftrightarrow\sqrt{a}\in\left\{4;2;5;1;7\right\}\)
\(\Leftrightarrow a\in\left\{16;4;25;1;49\right\}\)
Kết hợp ĐKXĐ, ta được: \(a\in\left\{1;16;25;49\right\}\)
Vậy: Để A nguyên thì \(a\in\left\{1;16;25;49\right\}\)
ĐK: \(a\ne-2\); \(a\in\mathbb{Z}\)
\(P=\dfrac{a-1}{a+2}=\dfrac{a+2-3}{a+2}=1-\dfrac{3}{a+2}\)
Để \(P\in\mathbb{Z}\) thì \(\dfrac{3}{a+2}\in\mathbb{Z}\)
\(\Rightarrow3⋮a+2\)
\(\Rightarrow a+2\inƯ\left(3\right)\)
\(\Rightarrow a+2\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow a\in\left\{-1;1;-3;-5\right\}\) (tmđk)