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a) P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\left(ĐKXĐ:a\ge0;a\ne4\right)\)
P = \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}-\dfrac{1}{\sqrt{a}-2}\)
P = \(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b) Ta có: P = \(\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\) = 1 - \(\dfrac{2}{\sqrt{a}-2}\)
Để \(P\in Z\) <=> 1 - \(\dfrac{2}{\sqrt{a}-2}\) \(\in Z\) <=> \(\sqrt{a}-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Ta có bảng sau:
| \(\sqrt{a}-2\) | 1 | -1 | 2 | -2 |
| \(\sqrt{a}\) | 3 | 1 | 4 | 0 |
| a | 9 (TM) | 1 (TM) | 16 (TM) | 0 (TM) |
Vậy để \(P\in Z\) thì \(a\in\left\{0;1;9;16\right\}\)
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 $a\in\left\{16;1;25;49\right\}$
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 \(a\in\left\{1;16;25;49\right\}\)
a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}:\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)^2}\)
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
ĐK: x>0,x\(\ne4\)
a) Ta thay x=\(\dfrac{1}{4}\) vào \(A=\dfrac{6}{x+2\sqrt{x}}=\dfrac{6}{\dfrac{1}{4}+2\sqrt{\dfrac{1}{4}}}=\dfrac{6}{\dfrac{1}{4}+2.\dfrac{1}{2}}=\dfrac{6}{\dfrac{1}{4}+1}=6:\left(\dfrac{1}{4}+1\right)=6:\dfrac{5}{4}=6.\dfrac{4}{5}=\dfrac{24}{5}=4,8\)B=\(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{4-x}\)
b) Ta có M=\(\dfrac{A}{B}=A\div B=\dfrac{6}{x+2\sqrt{x}}\div\dfrac{6}{4-x}=\dfrac{6}{x+2\sqrt{x}}.\dfrac{4-x}{6}=\dfrac{4-x}{x+2\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
Ta lại có M>1\(\Leftrightarrow\dfrac{2-\sqrt{x}}{\sqrt{x}}>1\Leftrightarrow2-\sqrt{x}>\sqrt{x}\Leftrightarrow2>2\sqrt{x}\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Kết hợp với ĐK
Vậy 0<x<1 thì M>1
c) Ta có M\(=\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{2}{\sqrt{x}}-1\)
Vậy để \(M\in Z\) thì \(\sqrt{x}\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}>0\)
Nên \(\sqrt{x}\in\left\{1;2\right\}\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)
Vậy x=1 thì M\(\in Z\)
Nguyễn Việt LâmTrầNguyễn Thị Khánh Như Trương NgọcThảo Vyn Trung NguyênBonkingsaint suppapong udomkaewkanjanaPhạm TiếnKHUÊ VŨMysterious PersonThiên Hàn
1) a) \(\sqrt{27}\) + \(\sqrt{75}\) - \(\sqrt{\dfrac{1}{3}}\) = \(3\sqrt{3}\) + \(5\sqrt{3}\) - \(\dfrac{\sqrt{3}}{3}\) = \(8\sqrt{3}\) - \(\dfrac{\sqrt{3}}{3}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) \(\sqrt{4+2\sqrt{3}}\) \(-\sqrt{4-2\sqrt{3}}\)
= \(\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\) \(-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}\)
= \(\sqrt{\left(\sqrt{3}+1\right)^2}\) \(-\sqrt{\left(\sqrt{3}-1\right)^2}\)
= \(\left(\sqrt{3}+1\right)\) \(-\left(\sqrt{3}-1\right)\)
= \(\sqrt{3}+1-\sqrt{3}+1\)
= 2
2) \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\) : \(\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\) : \(\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\left(\dfrac{a-1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right)\) : \(\left(\dfrac{\left(\sqrt{a}-1\right)+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right)\) : \(\left(\dfrac{\left(\sqrt{a}-1\right)+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) : \(\dfrac{2}{\sqrt{a}+1}\) = \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) . \(\dfrac{\sqrt{a}+1}{2}\) = \(\dfrac{\left(\sqrt{a}+1\right)^2}{2\sqrt{a}}\)
a: \(A=\dfrac{2\sqrt{a}-9-a+9+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)
b: Khi a=3+2 căn 2 thì
\(A=\dfrac{\sqrt{2}+1+1}{\sqrt{2}+1-3}=\dfrac{2+\sqrt{2}}{\sqrt{2}-2}=\dfrac{\sqrt{2}+1}{1-\sqrt{2}}=-3-2\sqrt{2}\)
c: Để A<1 thì A-1<0
=>căn a+1-căn a+3/căn a-3<0
=>căn a-3<0
=>0<=a<9
a,bn viết đúng đề xíu nhé \(\dfrac{\sqrt{a}+2}{\sqrt{a+3}}\) sửa \(\dfrac{\sqrt{a}+2}{\sqrt{a}+3}\)
đk: \(a\ge0,a\ne4\)
=>\(P=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{1}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)\(=\dfrac{\left(\sqrt{a}-4\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b, \(P=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}=1+\dfrac{-2}{\sqrt{a}-2}\) nguyên\(< =>\sqrt{a}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(=>a\in\left\{9;1;16;0\right\}\)(TM)