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a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)
\(=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\) khi
\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)
\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)
\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)
\(a,\) Rút gọn
\(A=\dfrac{3}{\sqrt{7}-2}+\sqrt{\left(\sqrt{7}-3\right)^2}\)
\(=\dfrac{3}{\sqrt{7}-2}+\left|\sqrt{7}-3\right|\)
\(=\dfrac{3}{\sqrt{7}-2}+3-\sqrt{7}\)
\(=\dfrac{3+\left(3-\sqrt{7}\right)\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=\dfrac{3+3\sqrt{7}-6-7+2\sqrt{7}}{\sqrt{7}-2}\)
\(=\dfrac{5\sqrt{7}-10}{\sqrt{7}-2}\)
\(=\dfrac{5\left(\sqrt{7}-2\right)}{\sqrt{7}-2}\)
\(=5\)
Vậy \(A=5\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{x-1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{x-\sqrt{x}}.\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}-1\)
Vậy \(B=\sqrt{x}-1\)
\(b,\) Để \(B< A\) thì \(\sqrt{x}-1< 5\)
\(\Leftrightarrow\sqrt{x}< 6\)
\(\Leftrightarrow x< 36\)
ĐKXĐ: \(x>0;x\ne9\)
\(P=\left(\dfrac{x+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+7-4\sqrt{x}-4+\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\right).\left(\dfrac{\sqrt{x}+6}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}.\dfrac{\left(\sqrt{x}+6\right)}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}+1}\)
b.
Ta có \(P=\dfrac{\sqrt{x}+1+5}{\sqrt{x}+1}=1+\dfrac{5}{\sqrt{x}+1}\)
Do \(\sqrt{x}+1>0\Rightarrow\dfrac{5}{\sqrt{x}+1}>0\Rightarrow P>1\)
\(P=\dfrac{6\left(\sqrt{x}+1\right)-5\sqrt{x}}{\sqrt{x}+1}=6-\dfrac{5\sqrt{x}}{\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}5\sqrt{x}>0\\\sqrt{x}+1>0\end{matrix}\right.\) ;\(\forall x>0\Rightarrow\dfrac{5\sqrt{x}}{\sqrt{x}+1}>0\)
\(\Rightarrow P< 6\Rightarrow1< P< 6\)
Mà P nguyên \(\Rightarrow P=\left\{2;3;4;5\right\}\)
- Để \(P=2\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=2\Rightarrow\sqrt{x}+6=2\sqrt{x}+2\Rightarrow x=16\)
- Để \(P=3\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=3\Rightarrow\sqrt{x}+6=3\sqrt{x}+3\Rightarrow\sqrt{x}=\dfrac{3}{2}\Rightarrow x=\dfrac{9}{4}\)
- Để \(P=4\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=4\Rightarrow\sqrt{x}+6=4\sqrt{x}+4\Rightarrow\sqrt{x}=\dfrac{2}{3}\Rightarrow x=\dfrac{4}{9}\)
- Để \(P=5\Rightarrow\dfrac{\sqrt{x}+6}{\sqrt{x}+1}=5\Rightarrow\sqrt{x}+6=5\sqrt{x}+5\Rightarrow\sqrt{x}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{16}\)
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
a: \(A=5\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-1\)
\(B=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: A=B
=>căn x=-căn x+1
=>căn x=1/2
=>x=1/4
a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)
\(=3-2=1\)
\(B=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b: B<A
=>B-1<0
=>\(\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}}< 0\)
=>-1/căn x<0
=>căn x>0
=>x>0 và x<>1
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)
a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)
\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)
Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)