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ĐKXĐ: a > 0
a/ \(K=\left[\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{1}{\sqrt{a}-1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\left[\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)
\(=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+3}\right]\) \(=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)
b/ Ta có: \(\sqrt{a}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
\(K=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}=\frac{\left(\sqrt{2}+2\right)\sqrt{2}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+4\right)}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}\left(\sqrt{2}+1\right)\left(2\sqrt{2}+1\right)}\)
\(=\frac{\sqrt{2}}{2\sqrt{2}+1}\)
c/ \(K< 0\Leftrightarrow\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)\(\Rightarrow\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)< 0\)
\(\Rightarrow\sqrt{a}-1< 0\) (vì \(\left(\sqrt{a}+1\right)^2>0\)) \(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)
Vậy \(0< a< 1\) thì K < 0
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:
\(\frac{a-1}{\sqrt{a}}>0\)
\(\Rightarrow a-1>0\)
\(\Rightarrow a>1\)
1/ a/ \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}-1+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(K=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(K=\frac{\sqrt{a}+1}{\sqrt{a}}:\frac{1}{\sqrt{a}-1}=\frac{\sqrt{a}+1}{\sqrt{a}}.\left(\sqrt{a}-1\right)\)
\(K=\frac{a-1}{\sqrt{a}}\)
b/ Với \(a=3+2\sqrt{2}\) => \(K=\frac{a-1}{\sqrt{a}}=\frac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}=\frac{2+2\sqrt{2}}{\sqrt{2+2\sqrt{2}+1}}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}=\frac{2\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\)
=> \(K=2\)
2/ Ta có: x3-y3=x-y)(x2+xy+y2)=(x-y)(x2-2xy+y2+3xy)=(x-y)[(x-y)2+3xy]=9
Thay x-y=3 vào ta được: 3(9+3xy)=9
<=> 3+xy=1 => xy=-2
Ta có hệ PT: \(\hept{\begin{cases}x-y=3\\xy=-2\end{cases}}\)=> \(\hept{\begin{cases}x=y+3\\xy=-2\end{cases}}\)
=> y(y+3)+2=0
<=> y2+3y+2=0
<=> y2+y+2y+2=0 <=> y(y+1)+2(y+1)=0 <=> (y+1)(y+2)=0
=> y1=-1 => x1=2
y2=-2 => x2=1
Đáp số: Các cặp x,y là: (2; -1) và (1; -2)
a. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\).\(\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
= \(\left[\left(\frac{\sqrt{a}}{2}\right)^2-2\frac{\sqrt{a}}{2}\frac{1}{2\sqrt{a}}+\left(\frac{1}{2\sqrt{a}}\right)^2\right]\).\(\left[\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{a-1}\right]\)
=\(\left(\frac{a}{4}-\frac{1}{2}+\frac{1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1\right)^2}{a-1}\cdot\frac{\left(\sqrt{a}+1\right)^2}{a-1}\right]\)
=\(\left(\frac{a^2}{4a}-\frac{2a}{4a}+\frac{1}{4a}\right)\).\(\left[\frac{\left[\left(\sqrt{a}-1\right)-\left(\sqrt{a}+1\right)\right]\cdot\left[\left(\sqrt{a}-1\right)+\left(\sqrt{a}+1\right)\right]}{a-1}\right]\)
=\(\left(\frac{a^2-2a+1}{4a}\right)\).\(\left[\frac{\left(\sqrt{a}-1-\sqrt{a}+1\right).\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right]\)
=\(\frac{\left(a-1\right)^2}{1}\).\(\frac{-4\sqrt{a}}{a-1}\)
=\(\frac{-\left(a-1\right)}{1}\)= - a + 1
hok tốt
đk: a>0; a khác 1
\(K=\left(\frac{\sqrt{a}.\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{a-1}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}\)
\(a=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\). thay a vào rồi tính thôi nha
b) \(K=\frac{a-1}{\sqrt{a}}\). ta thấy \(a>0\Rightarrow\sqrt{a}>0\) => K<0 <=> a-1<0 <=> a<1 => 0<a<1