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Bài 1 :
a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)
\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)
\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)
\(A=\sqrt{7}-\sqrt{28}\)
\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)
Vậy \(A=-\sqrt{7}\)
b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)
\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)
\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(B=a-b\)
Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)
_Minh ngụy_
Bài 2 :
a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))
Vậy \(x>1\)thì \(B>0\)
_Minh ngụy_
a, Với \(x>0;x\ne1\)
\(P=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\frac{x^2-2x+1}{4x}.\frac{-4\sqrt{x}}{x-1}=\frac{1-x}{\sqrt{x}}\)
Thay x = 4 => \(\sqrt{x}=2\)vào P ta được :
\(\frac{1-4}{2}=-\frac{3}{2}\)
c, Ta có : \(P< 0\Rightarrow\frac{1-x}{\sqrt{x}}< 0\Rightarrow1-x< 0\)vì \(\sqrt{x}>0\)
\(\Rightarrow-x< -1\Leftrightarrow x>1\)
#)Giải :
a) Câu trc của bn mk có giải rùi, thắc mắc vô Thống kê hđ của mk xem lại nhé !
b) Để \(P>0\Rightarrow\frac{x-1}{\sqrt{x}}>0\Rightarrow x-1>0\left(\sqrt{x}>0\right)\Rightarrow x>1\)
c) Bó tay @@
\(a,P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1}{\sqrt{x}}\)
Vậy với \(x>0;x\ne1\)thì \(P=\frac{x-1}{\sqrt{x}}\)
\(b,\)Để \(P>0\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\Leftrightarrow x-1>0\Leftrightarrow x>1\left(\sqrt{x}>0\right)\)
1)))))))
\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)
\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right)\div\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\div\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\times\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để A > 0
=> \(\frac{\sqrt{x}-1}{\sqrt{x}}>0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>0\end{cases}}\Leftrightarrow x>1\)
2. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}< 0\end{cases}}\)( dễ thấy trường hợp này không xảy ra :> )
Vậy với x > 1 thì A > 0
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
Trả lời:
a, \(B=\left(\frac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, \(B< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)
\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
Vì \(-\left(\sqrt{x}-1\right)^2< 0\) với mọi \(x>0;x\ne1\)
\(3\left(x+\sqrt{x}+1\right)>0\) với mọi \(x>0;x\ne1\)
\(\Rightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\) luôn đúng với mọi \(x>0;x\ne1\)
Vậy \(B< \frac{1}{3}\)
c, \(B=\frac{1}{2\sqrt{x}+1}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{1}{2\sqrt{x}+1}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=x+\sqrt{x}+1\)
\(\Leftrightarrow2x+\sqrt{x}=x+\sqrt{x}+1\)
\(\Leftrightarrow x=1\) (tm)
Vậy x = 1 là giá trị cần tìm