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a. ĐK \(\hept{\begin{cases}a\ge0\\a\ne4\\a\ne9\end{cases}}\)
P=\(\frac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{2\sqrt{a}-9-a+9+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)
b. P = \(\frac{\sqrt{a}+1}{\sqrt{a}-3}=1+\frac{4}{\sqrt{a}-3}\)
P nguyên \(\sqrt{a}-3\inƯ\left(4\right)\Rightarrow\sqrt{a}-3\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{a}\in\left\{1;2;4;5;7\right\}\Rightarrow a\in\left\{1;4;16;25;49\right\}\)
c. \(P< 1\Rightarrow P-1< 0\Rightarrow\frac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\Rightarrow\frac{4}{\sqrt{a}-3}< 0\)
\(\Rightarrow0\le a< 9\)và \(a\ne4\)
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a) \(ĐKXĐ:x\ne4;x\ne9\)
b) \(A=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{-\sqrt{x}+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
c) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) (ĐK: x thuộc Z)
\(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
\(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
x | 2 | \(\sqrt{2}\) | \(\sqrt{5}\) | \(\sqrt{1}\) | \(\sqrt{7}\) | \(\varnothing\) |
Vậy để A thuộc Z khi x = {2;\(\sqrt{2};\sqrt{5};\sqrt{1};\sqrt{7}\) }
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\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)
<=> \(\frac{5\left(a-b\sqrt{2}\right)}{a^2-2b^2}-\frac{4\left(a+b\sqrt{2}\right)}{a^2-2b^2}+18\sqrt{2}=3\) trục căn thức
<=> \(\frac{5a}{a^2-2b^2}-\frac{5b\sqrt{2}}{a^2-2b^2}-\frac{4a}{a^2-2b^2}-\frac{4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=3\)
Vì a; b nguyên => \(\hept{\begin{cases}\frac{5a}{a^2-2b^2}-\frac{4a}{a^2-2b^2}=3\\-\frac{5b\sqrt{2}}{a^2-2b^2}-\frac{4b\sqrt{2}}{a^2-2b^2}+18\sqrt{2}=0\end{cases}}\)
<=> \(\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{9b}{a^2-2b^2}=18\end{cases}}\)<=> \(\hept{\begin{cases}\frac{a}{a^2-2b^2}=3\\\frac{b}{a^2-2b^2}=2\end{cases}}\)
Với b = 0 => loại
Với b khác 0:
=> \(\frac{a}{b}=\frac{3}{2}\Leftrightarrow a=\frac{3}{2}b\)
=> \(\frac{b}{\frac{9}{4}b^2-2b^2}=2\)=> b = 2 => a = 3 thử lại thỏa mãn
Vậy a = 3 và b = 2.
\(\frac{5}{a+b\sqrt{2}}-\frac{4}{a-b\sqrt{2}}+18\sqrt{2}=3\)
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18\sqrt{2}\left(a^2-2b^2\right)=3\left(a^2-2b^2\right)\)
\(\Leftrightarrow5a-5b\sqrt{2}-4a-4b\sqrt{2}+18a^2\sqrt{2}-36b^2\sqrt{2}=3a^2-6b^2\)
\(\Leftrightarrow\left(18a^2-36b^2-9b\right)\sqrt{2}=3a^2-6b^2-a\)
-Nếu \(18a^2-36b^2-9b\ne0\Rightarrow\sqrt{2}=\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\)
Vì a,b nguyên nên \(\frac{3a^2-6b^2-a}{18a^2-36b^2-9b}\inℚ\Rightarrow\sqrt{2}\inℚ\)=> Vô lý vì \(\sqrt{2}\)là số vô tỷ
-Vậy ta có: \(18a^2-36b^2-9b=0\Rightarrow\hept{\begin{cases}18a^2-36b^2-9b=0\\3a^2-6b^2-a=0\end{cases}\Rightarrow\hept{\begin{cases}3a^2-6b^2=\frac{3}{2}b\\3a^2-6b^2=2\end{cases}}\Leftrightarrow a=\frac{3}{2}b}\)
Thay a=\(\frac{3}{2}b\)vào \(3a^2-6b^2-a=0\)
ta có \(3\cdot\frac{9}{4}b^2-6b^2-\frac{3}{2}b=0\Leftrightarrow27b^2-6b=0\Leftrightarrow3b\left(b-2\right)=0\)
Ta có b=0 (loại), b=2 (tm) => a=3
Vậy b=2; a=3
![](https://rs.olm.vn/images/avt/0.png?1311)
điều kiện \(x\ge0\)và x khác 1/4
Q= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{2x+7\sqrt{x}-4}\)
=\(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)
đề Q>1/2 thì \(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}>\frac{1}{2}\)
<=> \(8x+34\sqrt{x}+8>2x+7\sqrt{x}-4\)<=> \(6x+27\sqrt{x}+12>0\) với mọi x>=0
vậy Q>1/2 khi x>=0 và x khác 1/4