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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_
\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
a) ĐKXĐ :
\(\hept{\begin{cases}a\ge0\\a\ne4\end{cases}}\)
b) Với \(a\ge0\) và \(a\ne4\)
\(A=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\frac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\frac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)
Để A > 2
thì \(\frac{\sqrt{a}-4}{\sqrt{a}-2}>2\)
Ta có :
\(\frac{\sqrt{a}-4}{\sqrt{a}-2}-2\)
\(=\frac{\sqrt{a}-4-2\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)
\(=\frac{\sqrt{a}-4-2\sqrt{a}+4}{\sqrt{a}-2}\)
\(\)\(=\frac{-\sqrt{a}}{\sqrt{a}-2}\)
+) \(-\sqrt{a}< 0\forall a\) \(\Rightarrow a>0\)
+) \(\sqrt{a}-2< 0\) \(\Leftrightarrow a< 4\)
Vậy để A > 2 thì 0 < a < 4
c) Để A = 5
thì \(\frac{\sqrt{a}-4}{\sqrt{a}-2}=5\)
\(\frac{\left(\sqrt{a}-4\right)-5\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)}=0\)
\(\frac{\sqrt{a}-4-5\sqrt{a}+10}{\sqrt{a}-2}=0\)
\(\Rightarrow-4\sqrt{a}+6=0\)
\(\Rightarrow a=\frac{9}{4}\)( TMĐKXĐ )
Vậy để A = 5 thì a = 9/4
a, A xđ <=> \(\hept{\begin{cases}\sqrt{a}+3\ne0\\a+\sqrt{a}-6\ne0\\2-\sqrt{a}\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}a\ge0\\a\ne2\\a\ne4\end{cases}};a\ne-3\)-3
b, rút gọn: A=\(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)để A> 2 <=> \(\frac{\sqrt{a}-4}{\sqrt{a}-2}\)>2 <=> 1+\(\frac{-2}{\sqrt{a}-2}\)>2 <=> \(\frac{\sqrt{a}}{2-\sqrt{a}}\)>0
mà a\(\ge\)0 <=> \(\sqrt{a}\ge0\)=> \(2-\sqrt{a}\)>0 <=> a<4
kết hợp với điều kiện, ta được: \(0\le a< 4;a\ne2\)
c, để A = 5 thì \(\frac{-2}{\sqrt{a}-2}\)+1=5
<=> \(\frac{-2}{\sqrt{a}-2}\)=4
<=> \(a=\frac{9}{4}\)(t/m)
KL..............
1)đặt nhân tử chung quy đồng là xong
2)phân tích x+2cănx-3=(1-cănx)(3+cănx)
3)2a+căn a đặt căn a ra r rút gọn