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Ta có: \(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}}{4-\sqrt{a}}\)
a) ĐKXĐ: \(a\ne4;a\ne16;a\ge0\)
\(P=\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}-\dfrac{4\sqrt{a}}{\sqrt{a}-4}\)
\(P=\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{4\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(P=\dfrac{a+3\sqrt{a}+2\sqrt{a}+6-a+2\sqrt{a}+\sqrt{a}-2-4\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\)
\(P=\dfrac{4\sqrt{a}+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\)
\(P=\dfrac{4\sqrt{a}+4}{a-4}\)
b) Thay x=9 vào P ta có:
\(P=\dfrac{4\cdot\sqrt{9}+4}{9-4}=\dfrac{16}{5}\)
c) \(P< 0\) khi:
\(\dfrac{4\sqrt{x}+4}{a-4}< 0\)
Mà: \(4\sqrt{x}+4>0\)
\(\Rightarrow a-4< 0\)
\(\Rightarrow a< 4\)
kết hợp với Đk ta có:
\(0\le x< 4\)
a: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}+1}{3-\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}-9}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}+\dfrac{2\sqrt{a}+1}{\sqrt{a}-3}\)
\(=\dfrac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{2\sqrt{a}-9-a+9+2a-3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)
b: A<1
=>A-1<0
=>\(\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-1< 0\)
=>\(\dfrac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\)
=>\(\dfrac{4}{\sqrt{a}-3}< 0\)
=>căn a-3<0
=>0<=a<9 và a<>4
c: A là số nguyên
=>\(\sqrt{a}+1⋮\sqrt{a}-3\)
=>căn a-3+4 chia hết cho căn a-3
=>căn a-3 thuộc {1;-1;2;-2;4;-4}
mà a>=0 và a<>4; a<>9
nên a thuộc {16;25;1;49}
a: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}-1}{3-\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{2\sqrt{a}-9-a+9+2a-5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-3\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{\sqrt{a}-1}{\sqrt{a}-3}\)
b: A là số nguyên
=>\(\sqrt{a}-3+2⋮\sqrt{a}-3\)
=>\(\sqrt{a}-3\in\left\{1;-1;2;-2\right\}\)
=>a thuộc {16;25;1}
a, \(B=\frac{\sqrt{a}+3}{2\sqrt{a}-6}-\frac{3-\sqrt{a}}{2\sqrt{a}+6}=\frac{\left(2\sqrt{a}+6\right)\left(\sqrt{a}+3\right)+\left(2\sqrt{a}-6\right)\left(\sqrt{a}-3\right)}{4a-36}\)
\(=\frac{2a+12\sqrt{a}+18+2a-12\sqrt{a}+18}{4a-36}=\frac{4a+36}{4a-36}=\frac{a+9}{a-9}\)
b, Ta có : \(B>1\Rightarrow\frac{a+9}{a-9}>1\Leftrightarrow\frac{a+9}{a-9}-1>0\)
\(\Leftrightarrow\frac{a+9-a+9}{a-9}>0\Leftrightarrow\frac{18}{a-9}>0\Rightarrow a-9>0\Leftrightarrow a>9\)vì 18 > 0
\(B< 1\Rightarrow\frac{a+9}{a-9}< 1\Leftrightarrow\frac{a+9}{a-9}-1< 0\)
\(\Leftrightarrow\frac{a+9-a+9}{a-9}< 0\Leftrightarrow\frac{18}{a-9}< 0\Rightarrow a-9< 0\Leftrightarrow a< 9\)vì 18 > 0
c, Ta có : \(B=4\Rightarrow\frac{a+9}{a-9}=4\Rightarrow a+9=4a-36\Leftrightarrow3a=45\Leftrightarrow a=15\)
Vậy a = 15 thì B = 4
a: ĐKXĐ: x>=0; x<>1
b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)
d: căn x+2>=2
=>A<=1/2
Dấu = xảy ra khi x=0