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Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a: \(P=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2+x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{3x+\sqrt{x}+3-x+\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{2\cdot\left(3-2\sqrt{2}\right)+2\left(\sqrt{2}-1\right)+2}{\sqrt{2}-1}\)
\(=\dfrac{6-4\sqrt{2}+2\sqrt{2}-2+2}{\sqrt{2}-1}=\dfrac{6-2\sqrt{2}}{\sqrt{2}-1}=4\sqrt{2}+2\)
a) P = \(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}-\frac{\sqrt{x}-1}{x+\sqrt{x}}\right)\).
P = \(\frac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}-1\right)}\)
P = \(\frac{x-1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1-x+\sqrt{x}}\)
P = \(\frac{x-1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
P = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
P = \(\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
P = \(x-1\).
b) P = \(\frac{9}{2}\).
⇔ \(x-1=\frac{9}{2}\)
⇔ \(x=\frac{11}{2}\).
Vậy \(x=\frac{11}{2}\)thì P = \(\frac{9}{2}\).
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
a: \(S=\dfrac{x+1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(x+1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
b: Khi \(x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\) vào S, ta được:
\(S=\dfrac{\left(4-2\sqrt{3}+1\right)\left(\sqrt{3}-1+1\right)}{\left(\sqrt{3}-1\right)\left(4-2\sqrt{3}-1\right)}\)
\(=\dfrac{\left(5-2\sqrt{3}\right)\cdot\sqrt{3}}{\left(\sqrt{3}-1\right)\left(3-2\sqrt{3}\right)}\)
a: \(=\dfrac{\sqrt{x}+1-2x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2}{\sqrt{x}+1}\)
b: Để D=-1/2 thì \(\sqrt{x}+1=4\)
=>x=9
a/ \(P=\dfrac{1}{\sqrt{x}+1}+\dfrac{x}{\sqrt{x}-x}=\dfrac{\sqrt{x}-x+x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-x\right)}\)
\(=\dfrac{\sqrt{x}-x+x\sqrt{x}+x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-x\right)}\)\(=\dfrac{\sqrt{x}+x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-x\right)}\)
\(=\dfrac{\sqrt{x}\left(x+1\right)}{\left(\sqrt{x}+1\right)\sqrt{x}\left(1-\sqrt{x}\right)}\)\(=\dfrac{x+1}{1-x}\)
b/ thay x = \(\dfrac{1}{\sqrt{2}}\) vào P:
\(P=\dfrac{\dfrac{1}{\sqrt{2}}+1}{1-\dfrac{1}{\sqrt{2}}}=3+2\sqrt{2}\)