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1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
ĐKXĐ: x \(\ge\)0; x \(\ne\)1
a) P = \(\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
P = \(\left(\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{5}{x+2\sqrt{x}-\sqrt{x}-2}\right):\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)
P = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b) P = \(\frac{1}{\sqrt{x}}\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)
=> \(\sqrt{x}\left(2\sqrt{x}+1\right)-\sqrt{x}-1=0\)
<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)
<=> \(x=\frac{1}{2}\)(tm)
c)Với đk: x \(\ge\)0 và x \(\ne\)1
\(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))
<=> \(x-1-2\sqrt{x-1}+1=0\)
<=> \(\left(\sqrt{x-1}-1\right)^2=0\)
<=> \(\sqrt{x-1}-1=0\)
<=> \(\sqrt{x-1}=1\)
<=> \(\left(\sqrt{x-1}\right)^2=1\)
<=> \(\left|x-1\right|=1\)
<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
Với x = 2 => P = \(\frac{2\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{4-2\sqrt{2}+\sqrt{2}-1}{2-1}=3-\sqrt{2}\)
a) P = \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)(sửa lại)
b) \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\) => \(2x-\sqrt{x}-\sqrt{x}-1=0\)
<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)
<=> \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)
a, \(A=\left(\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}\right):\frac{-\sqrt{x}}{x-2\sqrt{x}}\)
\(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\frac{-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(A=\frac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}}\)
\(A=\frac{4}{\sqrt{x}+2}\)
b, \(A=\frac{4}{\sqrt{x}+2}=\frac{2}{3}\)
=> 2cawn x + 4 = 12
=> 2.căn x = 8
=> căn x = 4
=> x = 16 (thỏa mãn)
c, có A = 4/ căn x + 2 và B = 1/căn x - 2
=> A.B = 4/x - 4
mà AB nguyên
=> 4 ⋮ x - 4
=> x - 4 thuộc Ư(4)
=> x - 4 thuộc {-1;1;-2;2;-4;4}
=> x thuộc {3;5;2;6;0;8} mà x > 0 và x khác 4
=> x thuộc {3;5;2;6;8}
d, giống c thôi
1. \(N=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\right):\frac{\sqrt{x}-3}{2\sqrt{x}-x}\)
\(N=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(N=\left(\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(N=\left(\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\frac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(N=\left(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right).\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(N=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\)
\(N=\frac{4x}{x-3}\)
Vậy \(N=\frac{4x}{x-3}\)với \(x>0,x\ne4,x\ne9\)
2.Với \(x>0,x\ne4,x\ne9\)
Ta có \(N< 0\)\(\Leftrightarrow\frac{4x}{x-3}< 0\)\(\Leftrightarrow x-3< 0\)(Vì \(x>0\Leftrightarrow4x>0\)\(với\forall x\))\(\Leftrightarrow x< 3\)
Vậy ..........
3. Với \(x>0,x\ne4,x\ne9\)
Ta có \(\left|N\right|=1\Leftrightarrow\left|\frac{4x}{x-3}\right|=1\Leftrightarrow\orbr{\begin{cases}\frac{4x}{x-3}=1\\\frac{4x}{x-3}=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}4x=3-x\\4x=x-3\end{cases}}\)\(\orbr{\begin{cases}x=\frac{3}{5} \left(N\right)\\x=-1\left(N\right)\end{cases}}\)
Vậy ...............
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_
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 \(x=\frac{1}{1-2\sqrt{5}}\)
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 :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)