\(\dfrac{\sqrt{\left(x-5\right)^2}}{\left(4-x\right)^2}-\dfrac{x^2-25}{x-4}\)
tại x = 3
\(\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3\)
\(\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}\)
\(\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4\)
\(\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}\)
\(\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6\)
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
a, \(\dfrac{\sqrt[]{7-2\sqrt[]{6}}}{\sqrt[]{6}-1}\)
b, 2.|x+y|.\(\sqrt[]{\dfrac{1}{x^2+2xy+y^2}}\) (x+y>0)
c, \(\dfrac{\left(x-5\right)^4}{\left(4-x\right)^2}\)-\(\dfrac{x^2-25}{x-4}\)(x<4)
Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
4) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
5) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
6) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp mình với, cần gấp ạ
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
Rút gọn:
1) \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
2) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
3) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2003}{x}\)
4) \(A=\left(\dfrac{5\sqrt{x}}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\left(2-\sqrt{x}\right)\)
5) \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
Giúp vs ạ
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
3: Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{x-1}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
1. Rút gọn biểu thức
\(\sqrt{\dfrac{4}{3}}+\sqrt{12}-\dfrac{4}{3}\sqrt{\dfrac{3}{4}}\)
2. Đưa thừa số vào trong dấu căn :
a. \(\left(2-a\right)\sqrt{\dfrac{2a}{a-2}}\) với a lớn hơn 2
b. với 0 bé hơn x, x bé hơn 5. \(\left(x-5\right)\sqrt{\dfrac{x}{25-x^2}}\)
c. Với 0 bé hơn a, a bé hơn b \(\left(a-b\right)\)\(\sqrt{\dfrac{3a}{b^2-a^2}}\)
\(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\right)\)
\(B=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{x+2}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
a) Rút gọn A & B
b) Tìm x để B > 0
c) Tính B khi \(\left|1-x\right|=0\)
tìm x để các biểu thức sau có nghĩa:
a)\(\sqrt{\left(x-2\right)}\)+\(\dfrac{1}{x-5}\) b)\(\sqrt{\left(2x-6\right)\left(7-x\right)}\) c)\(\sqrt{4x^2-25}\)
d)\(\dfrac{2}{x^2-9}\)-\(\sqrt{5-2x}\) e)\(\dfrac{x}{x^2-4}\)+\(\sqrt{x-2}\)
a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)
b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:
\(\left(x-3\right)\left(7-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow3\le x\le7\)
c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:
\(\left(2x-5\right)\left(2x+5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x>2\)
Rút gọn biểu thức
a) A=\(2\sqrt{\left(2-\sqrt{5}\right)^2}-\dfrac{8}{3-\sqrt{5}}\)
b) B= \(\left(\dfrac{2\sqrt{x}}{x-4}-\dfrac{1}{\sqrt{x}+2}\right):\left(1+\dfrac{2}{\sqrt{x}-2}\right)\) Với x>0, x khác 4
\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)
\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)
\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)