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a, Ta có: H = \(\frac{2x.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}\) + \(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\) - \(\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
= \(\frac{2x}{x-1}+\frac{\sqrt{x}-1}{x-1}-\frac{\sqrt{x}+1}{x-1}\)
= \(\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{x-1}\)
= \(\frac{2x-2}{x-1}\)
= 2
b, Ta có: \(\sqrt{x}\) < H <=> \(\sqrt{x}\) < 2
<=> x < 4
Vậy x = 4 thì \(\sqrt{x}\) < H
\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)
\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)
\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)
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 \(A>A^2\)
a) - ĐKXĐ: x\(\ge\)0 , x\(\ne1\)
- Với x\(\in\)ĐKXĐ. Rút gọn bt H ta được:
H= \(\frac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{2x}{x-1}+\frac{\sqrt{x}-1}{x-1}-\frac{\sqrt{x}+1}{x-1}\)=\(\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{x-1}=\frac{2x-2}{x-1}=\frac{2\left(x-1\right)}{x-1}=2\)
Vậy với x\(\ge\)0 , x\(\ne1\)thì H=2
b)Với x\(\ge0\), x\(\ne1\). Để \(\sqrt{x}H< 0\)\(\Leftrightarrow\) \(2\sqrt{x}< 0\)\(\Leftrightarrow\)\(\sqrt{x}< 0\) (vì 2>0)
\(\Leftrightarrow\)x<0 (vô lý)
Vậy không tồn tại giá trị của x để \(\sqrt{x}H< 0\)
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 \(x=\frac{1}{1-2\sqrt{5}}\)
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(H=\frac{2x^2+2x}{x^2-1}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow H=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow H=\frac{2x}{x-1}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow H=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow H=\frac{2x-2}{x-1}\)
\(\Leftrightarrow H=2\)
b) Để \(\sqrt{x}< H\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
Mà \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}0\le x< 1\\1< x< 4\end{cases}}\)
p/s : vì đề bài không yêu cầu \(x\)nguyên nên mình làm như vậy !