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N
1 tháng 1 2019
a)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)
\(=1\)
b)ĐK: \(x>0,x\ne9\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
30 tháng 7 2018
4 , Ta có :
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x-9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3\left(x-3\right)}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x+9}{x-9}\)
\(=\dfrac{3\sqrt{x}+9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}-3}\)
30 tháng 7 2018
2 , Ta có :
\(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x\sqrt{x}-x-\sqrt{x}+1}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
26 tháng 5 2022
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
18 tháng 9 2018
a) điều kiện xác định : \(x\ge1\)
ta có : \(\sqrt{\dfrac{x-1}{4}}-3=\sqrt{\dfrac{4x-4}{9}}\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-3=\dfrac{2}{3}\sqrt{x-1}\)
\(\Leftrightarrow\dfrac{1}{6}\sqrt{x-1}=-3\left(vôlí\right)\) vậy phương trình vô nghiệm
b) điều kiện xác định \(x\ge3\)
ta có : \(\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}=x-3\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}=x-3\) \(\Leftrightarrow\left|x-2\right|+\left|x+3\right|=x-3\)
\(\Leftrightarrow x-2+x+3=x-3\Leftrightarrow x=-4\left(L\right)\) vậy phương trình vô nghiệm
c) điều kiện xác định : \(\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\)
ta có : \(\sqrt{\dfrac{2x-3}{x-1}}=2\) \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\) vậy \(x=\dfrac{1}{2}\)
1 tháng 9 2018
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
1 tháng 9 2018
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
PM
15 tháng 7 2018
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 \(\left\{{}\begin{matrix}0\le x< 25\\x\ne9\end{matrix}\right.\)
5 tháng 9 2022
a: \(=\dfrac{1}{\sqrt{x-1}+1}+\dfrac{1}{\sqrt{x-1}-1}\)
\(=\dfrac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}=\dfrac{2\sqrt{x-1}}{x-2}\)
b: \(=\dfrac{1}{2\sqrt{3}+\sqrt{5}+2}-\dfrac{1}{2\sqrt{3}-\sqrt{5}+2}\)
\(=\dfrac{1}{\left(2\sqrt{3}+2\right)+\sqrt{5}}-\dfrac{1}{\left(2\sqrt{3}+2\right)-\sqrt{5}}\)
\(=\dfrac{2\sqrt{3}+2-\sqrt{5}-2\sqrt{3}-2-\sqrt{5}}{11+8\sqrt{3}}\)
\(=\dfrac{-2\sqrt{5}}{11+8\sqrt{3}}=\dfrac{\sqrt{5}\left(22-16\sqrt{3}\right)}{71}\)
CW
27 tháng 6 2018
a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)
\(=\sqrt{1+\sqrt{2}+\sqrt{3}-\left(\sqrt{3}+\sqrt{2}\right)}=1\)
b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)
\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)
Th1: x-3 < 0
\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)
Th2: x-3 > 0
\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)
c)
Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\dfrac{\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
Th1: \(x-2\ge0\Leftrightarrow x\ge2\)
\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)
Th2: \(x-2\le0\Leftrightarrow x\le2\)
kết hợp với đk, ta được: 1 \< x \< 2
\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)
d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
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Tìm điều kiện để biểu thức có nghĩa nha mọi người
Đặt \(A=\sqrt{x^2-9}+\dfrac{1}{\sqrt{x^2-9}}\)
A có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x^2-9}\ge0\\\dfrac{1}{\sqrt{x^2-9}}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\ge9\\x^2>9\end{matrix}\right.\)\(\Leftrightarrow x^2>9\)
\(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\)
Chúc học tốt nhé :))