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Lời giải:
a) ĐK: \(a>0; a\neq 1\)
\(K=\left(\frac{a}{\sqrt{a}(\sqrt{a}-1)}-\frac{1}{\sqrt{a}(\sqrt{a}-1)}\right): \left(\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}+\frac{2}{(\sqrt{a}-1)(\sqrt{a}+1)}\right)\)
\(=\frac{a-1}{\sqrt{a}(\sqrt{a}-1)}: \frac{\sqrt{a}+1+2}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(=\frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}(\sqrt{a}-1)}. \frac{(\sqrt{a}-1)(\sqrt{a}+1)}{\sqrt{a}+3}\)
\(=\frac{(\sqrt{a}+1)^2(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}+3)}\)
b) \(a=3+2\sqrt{a}\Leftrightarrow a-2\sqrt{a}-3=0\)
\(\Leftrightarrow (\sqrt{a}-3)(\sqrt{a}+1)=0\)
\(\Rightarrow \sqrt{a}=3\)
Khi đó: \(K=\frac{(3+1)^2(3-1)}{3.(3+3)}=\frac{16}{9}\)
c) Để \(K< 0\Leftrightarrow \frac{(\sqrt{a}+1)^2(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}+3)}< 0\)
Mà \(\frac{(\sqrt{a}+1)^2}{\sqrt{a}(\sqrt{a}+3)}>0, \forall a> 0; a\neq 1\), do đó \(\sqrt{a}-1< 0\Leftrightarrow 0< a< 1\)
Vậy .........
Lời giải:
ĐK: \(a>0; a\neq 1\)
a) \(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right): \left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(B=\left(\frac{a}{a-\sqrt{a}}-\frac{1}{a-\sqrt{a}}\right): \left(\frac{\sqrt{a}-1}{(\sqrt{a}+1)(\sqrt{a}-1)}+\frac{2}{a-1}\right)\)
\(=\frac{a-1}{a-\sqrt{a}}:\left(\frac{\sqrt{a}-1}{a-1}+\frac{2}{a-1}\right)\)
\(=\frac{a-1}{a-\sqrt{a}}: \frac{\sqrt{a}+1}{a-1}=\frac{a-1}{a-\sqrt{a}}.\frac{a-1}{\sqrt{a}+1}=\frac{(a-1)^2}{\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}=\frac{(a-1)^2}{\sqrt{a}(a-1)}=\frac{a-1}{\sqrt{a}}\)
b) Ta có:
\(a=3+2\sqrt{2}=2+1+2\sqrt{2}=(\sqrt{2}+1)^2\)
\(\Rightarrow K=\frac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\frac{2+2\sqrt{2}}{\sqrt{2}+1}=\frac{2(1+\sqrt{2})}{\sqrt{2}+1}=2\)
c) \(K< 0\leftrightarrow \frac{a-1}{\sqrt{a}}< 0\Leftrightarrow a-1< 0\) (do \(\sqrt{a}>0\))
\(\Leftrightarrow a< 1\)
Vậy \(0< a< 1\)
Nhật Hạ : bạn ghi trên đề bài mà.
Thực ra nó chỉ là tên biểu thức nên không quan trọng.
a: \(K=\dfrac{a-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1+2}{a-1}\)
\(=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}+3}\)
\(=\dfrac{a\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
c: Vì \(\sqrt{a}+3>=3>0;\sqrt{a}>0;a\sqrt{a}+1>0\)
nên K>0 với mọi a thỏa mãn ĐKXĐ
=>Không có giá trị nào của a để K<0
Lời giải:
ĐK: \(x>0; x\neq 4\)
Có: \(K=\left(\frac{4\sqrt{x}(2-\sqrt{x})}{(2+\sqrt{x})(2-\sqrt{x})}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right)\)
\(=\frac{8\sqrt{x}-4x+8x}{(2+\sqrt{x})(2-\sqrt{x})}: \frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)
\(=\frac{8\sqrt{x}+4x}{(2+\sqrt{x})(2-\sqrt{x})}.\frac{\sqrt{x}(\sqrt{x}-2)}{-\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}(2+\sqrt{x})}{2+\sqrt{x}}. \frac{-\sqrt{x}}{3-\sqrt{x}}=\frac{-4\sqrt{x}.\sqrt{x}}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)
b)
\(K=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1\Rightarrow 4x=-(\sqrt{x}-3)\)
\(\Leftrightarrow 4x+\sqrt{x}-3=0\)
\(\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0\)
Vì \(\sqrt{x}+1>0\Rightarrow 4\sqrt{x}-3=0\Rightarrow x=\frac{9}{16}\)
c) \(m(\sqrt{x}-3)K>x+1\)
\(\Leftrightarrow m. (\sqrt{x}-3).\frac{4x}{\sqrt{x}-3}>x+1\)
\(\Leftrightarrow m> \frac{x+1}{4x}\)
\(\Leftrightarrow m> max(\frac{4x}{x+1}), \forall x< 9\)
Với đk đã cho thì ta thấy \(\frac{4x}{x+1}\) có min thôi.
P/s gọi a = x cho dễ viết nhé
a, Với \(x\ge0;x\ne1;x\ne4\)
\(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
chỗ này mình nghĩ ko phải trục căn thức đâu ha :D
b, Ta có P > 1/6 hay \(\frac{\sqrt{x}-2}{3\sqrt{x}}>\frac{1}{6}\Leftrightarrow\frac{\sqrt[]{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)
\(\Leftrightarrow\frac{6\sqrt{x}-12-3\sqrt{x}}{18\sqrt{x}}>0\Leftrightarrow\frac{3\sqrt{x}-12}{18\sqrt{x}}>0\)
\(\Leftrightarrow3\sqrt{x}-12>0\)( vì \(18\sqrt{x}>0\))
\(\Leftrightarrow3\sqrt{x}>12\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
Vậy \(x>16\)
cho mình hỏi đề có sai ko ? \(P< \frac{1}{6}\)mình nghĩ sẽ hợp lí hơn
んuﾘ ｲ hãy thuận theo ý thầy :)) và nhớ chú ý đến ĐKXĐ
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\left(\frac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\times\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Để P > 1/6 thì \(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
<=> \(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
<=> \(\frac{2\sqrt{a}-4}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\)
<=> \(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Dễ thấy \(6\sqrt{a}>0\forall x>0\)
=> \(\sqrt{a}-4>0\)<=> \(\sqrt{a}>4\)<=> \(a>16\)
Vậy với a > 16 thì P > 1/6
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a-1}\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\frac{\sqrt{a}+1}{\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}}\cdot\sqrt{a}-1\)
\(=\frac{a-1}{\sqrt{a}}\)
b) thay \(a=3+2\sqrt{2}\) vào bt K được:
\(\frac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}\) \(=\frac{2+2\sqrt{2}}{\sqrt{2+2\sqrt{2}+1}}\) \(=\frac{2\left(1+\sqrt{2}\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}\) \(=\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\) \(=2\)
c) để K>0 thì:
\(\frac{a-1}{\sqrt{a}}>0\)
\(\Rightarrow a-1>0\)
\(\Rightarrow a>1\)
a: \(K=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{a-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\)
b: Thay \(a=3+2\sqrt{2}\) vào K, ta được:
\(K=\dfrac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)
c: Để K<0 thì a-1<0
hay 0<a<1