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a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+\left|x-3\right|\)
\(=x+3-\left(x-3\right)\)
\(=x+3-x+3\)
\(=6\)
b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)
\(=\sqrt{\left(x+2\right)^2}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
\(=x+2-\left(-x\right)\)
\(=x+2+x\)
\(=2x+2=2\left(x+1\right)\)
c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)
\(=\frac{\sqrt{\left(x-1\right)^2}}{x-1}\)
\(=\frac{\left|x-1\right|}{x-1}\)
\(=\frac{x-1}{x-1}=1\)
d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)
\(=\left|x-2\right|+\frac{\sqrt{\left(x-2\right)^2}}{x-2}\)
\(=\left|x-2\right|+\frac{\left|x-2\right|}{x-2}\)
\(=\left|x-2\right|+\frac{-\left(x-2\right)}{x-2}\)
\(=\left|x-2\right|-1\)
\(=-\left(x-2\right)-1\)
\(=-x+2-1\)
\(=-x+1=-\left(x-1\right)\)
\(B=\sqrt{x+\sqrt{x^2-1}}-\sqrt{x-\sqrt{x^2-1}}\)
\(B^2=x+\sqrt{x^2-1}+x-\sqrt{x^2-1}-2\sqrt{\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)}\)
\(B^2=2x-2\sqrt{x^2-x^2+1}\)
\(B^2=2x-2\)
\(\Rightarrow B=\sqrt{2x-2}\)
\(C=\sqrt{x+2\sqrt{x-1}}-\sqrt{x-1}\left(ĐK:x\ge1\right)\)
\(C=\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{x-1}\)
\(C=\sqrt{x-1}+1-\sqrt{x-1}=1\)
Bài 1.
\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\div\frac{x}{x-\sqrt{x}}\)với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
a) \(B=\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\left(\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\div\frac{x}{x-\sqrt{x}}\)
\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\div\frac{x}{x-\sqrt{x}}\)
\(B=\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x}\)
\(B=\frac{4\sqrt{x}\cdot\sqrt{x}}{\left(\sqrt{x}+1\right)x}=\frac{4x}{\left(\sqrt{x}+1\right)x}=\frac{4}{\sqrt{x}+1}\)
b) Để B > 1
=> \(\frac{4}{\sqrt{x}+1}>0\)( với \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\))
Vì 4 > 0
=> \(\sqrt{x}+1>0\)
<=> \(\sqrt{x}>-1\)( luôn luôn đúng \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)) ( theo ĐKXĐ )
Vậy \(\forall\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)thì B > 1
Chưa chắc lắm ... Còn câu 2 thì tí nữa mình làm cho
Bài 2.
\(A=2\sqrt{5}-1\)
\(B=\frac{2}{x-1}\cdot\sqrt{\frac{x^2-2x+1}{4x^2}}\)( x > 0 )
a) \(B=\frac{2}{x-1}\cdot\frac{\sqrt{x^2-2x+1}}{\sqrt{4x^2}}\)
\(B=\frac{2}{x-1}\cdot\frac{\sqrt{\left(x-1\right)^2}}{\sqrt{\left(2x\right)^2}}\)
\(B=\frac{2}{x-1}\cdot\frac{\left|x-1\right|}{\left|2x\right|}\)
\(B=\frac{2}{x-1}\cdot\frac{x-1}{2x}=\frac{1}{x}\)( vì x > 0 )
b) Để A + B = 0
=> \(\left(2\sqrt{5}-1\right)+\frac{1}{x}=0\)( ĐKXĐ : \(x\ne0\))
<=> \(\frac{1}{x}=-\left(2\sqrt{5}-1\right)\)
<=> \(\frac{1}{x}=1-2\sqrt{5}\)
<=> \(x\times\left(1-2\sqrt{5}\right)=1\)
<=> \(x=\frac{1}{1-2\sqrt{5}}\)( tmđk )
Vậy \(x=\frac{1}{1-2\sqrt{5}}\)
a) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}=4x-\sqrt{8}+x=5x-\sqrt{8}\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)
\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)
b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)
\(2\sqrt{x}-1>0\);\(4x>0\)
\(\Rightarrow x>0\)thì \(A>A^2\)