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Bài 2:
\(\Leftrightarrow3\sqrt{x+5}-2\sqrt{x+5}=7\)
\(\Leftrightarrow\sqrt{x+5}=7\)
=>x+5=25
hay x=18
\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
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
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
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..............