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a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(=\dfrac{-6}{\sqrt{x}+3}\)
b: Để A<-1/2 thì A+1/2<0
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow-12+\sqrt{x}+3< 0\)
=>0<x<81 và x<>9
a: \(A=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}+\sqrt{5}+3-3-\sqrt{5}=1\)
b: \(B=\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{x-9}=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)
Để B>1 thì \(\dfrac{-4-\sqrt{x}+3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay 0<x<9
a: \(=-4+2\sqrt{5}-\sqrt{5}+2+\sqrt{5}=2\sqrt{5}-2\)
b: \(B=\dfrac{2\sqrt{x}+4+6\sqrt{x}-3-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{6\sqrt{x}+4}\)
\(=\dfrac{\left(6\sqrt{x}+1\right)\cdot\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(6\sqrt{x}+4\right)}\)
a/ \(\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}=\dfrac{x+3+2\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{x-9}=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{x-9}=\dfrac{x-\sqrt{x}-6}{x-9}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)}{x-9}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
b/ \(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\)
P/s: câu b đề sai phải không bạn, mk nghĩ ngoài dấu ngoặc là phép chia thì đúng hơn
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:
Sửa đề: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(C=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=-\dfrac{3\sqrt{x}}{2\sqrt{x}+4}\)
b: Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)
b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)
a)\(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(A=\dfrac{x\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(A=\dfrac{x\sqrt{x}-3-\left(2\sqrt{x}+6\right)-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{x\sqrt{x}-3-2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(A=\dfrac{-2\sqrt{x}-12}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
1, \(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(x-4\sqrt{x}-9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
2, Để P = 3 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-3}=3\Rightarrow3\sqrt{x}-9=\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}-9=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{9}{2}\Leftrightarrow x=\dfrac{81}{4}\)(thỏa mãn)
3, \(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{\sqrt{x}+5}{3-\sqrt{x}}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)
để \(\left|M\right|< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}+5}< \dfrac{1}{2}\) \(\Leftrightarrow2\sqrt{x}< \sqrt{x}+5\)
\(\Leftrightarrow\sqrt{x}< 5\)
\(\Leftrightarrow0\le x< 25\)
Kết hợp ĐK ta có \(\left\{{}\begin{matrix}0\le x< 25\\x\ne9\end{matrix}\right.\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
e cảm ơn ạ :33