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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
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.
\(P=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(P=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)}\)
\(P=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
b/
\(a=2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(a=\sqrt{3-\sqrt{5}}\left(6+2\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\)
\(a=\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)\)
\(a=\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2\)
\(a=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)
\(\Rightarrow P=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{16}+1}{\sqrt{16}}=\dfrac{4+1}{4}=\dfrac{5}{4}\)
1: \(A=\dfrac{a+1-2\sqrt{a}}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(=\sqrt{a}+1\)
2: Khi \(a=2010-2\sqrt{2009}\) thì \(A=\sqrt{2009}-1+1=\sqrt{2009}\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
a)\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\):\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a-1}\right)}\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\left(a-1\right)\left(a-4\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) ĐKXĐ: \(x>0\) \(a\ne4\) \(a\ne1\)
b) \(Q>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\) (Kết hợp ĐKXĐ \(a>0\))
\(\Leftrightarrow\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
\(\Leftrightarrow a>4\) (Thỏa mãn ĐKXĐ)
Vậy \(a>4\) thì \(Q>0\)
____♫ Chúc bạn học tốt ♫____
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 .........