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Bài 2:
a: ĐKXĐ: 2/3x-1/5>=0
=>2/3x>=1/5
hay x>=3/10
b: ĐKXĐ: \(\dfrac{x+1}{2x-3}>=0\)
=>2x-3>0 hoặc x+1<=0
=>x>3/2 hoặc x<=-1
c: ĐKXĐ: \(\left\{{}\begin{matrix}3x-5>=0\\x-4>0\end{matrix}\right.\Leftrightarrow x>4\)
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
a: ĐKXĐ: x>0; x<>1
\(M=\dfrac{x-\sqrt{x}-x-\sqrt{x}-1}{x-1}\cdot\dfrac{x}{2\sqrt{x}+1}\)
\(=\dfrac{-\left(2\sqrt{x}+1\right)}{x-1}\cdot\dfrac{x}{2\sqrt{x}+1}=\dfrac{-x}{x-1}\)
b: Khi \(x=\dfrac{\sqrt{3}-1}{2}\) thì \(M=\dfrac{-\sqrt{3}+1}{2}:\dfrac{-\sqrt{3}+1-2}{2}\)
\(=\dfrac{-\sqrt{3}+1}{-1-\sqrt{3}}=2-\sqrt{3}\)
a) điều kiện : \(x>0;x\ne4\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\) \(\left(x>0\right)\)
thay vào P ta có \(P=\dfrac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}+1-2\right)}=\dfrac{\sqrt{3}+3}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}+3}{2}\)
\(P>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)
ta có : \(\sqrt{x}+2>0\) và \(\sqrt{x}>0\) \(\left(x>0\right)\)
\(\Rightarrow p>0\) thì \(\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)
vậy \(x>4\) thì P > 0
câu : a ; b ; c đầy đủ rồi nha quênh gi câu : a ; b ; c 
a/ Đkxđ: x\(\ge\)0 x\(\ne\)4
=\(\frac{3\left(\sqrt{x}+2\right)+2\left(\sqrt{x}-2\right)+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{5\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\frac{5}{\sqrt{x}-2}\)
b/ Với x\(\ge\)0 vã\(\ne\)4
Để M\(\in\)Z \(\Leftrightarrow\) \(\frac{5}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\) \(\sqrt{x}-2\inƯ\left(5\right)\)
\(\begin{cases}\sqrt{x}-2=5\\\sqrt{x}-2=-5\\\sqrt{x}-2=1\\\sqrt{x}-2=-1\end{cases}\Rightarrow\begin{cases}x=49\left(tmĐKXĐ\right)\\KhongcogiatriTm\\x=9\left(tmĐKXĐ\right)\\x=1\left(tmĐKXĐ\right)\end{cases}\)
Vậy để M\(\in\)Z thì x=.....
c/ Với...
Để M<2 thì \(\frac{5}{\sqrt{x}-2}< 2\Rightarrow\frac{5-2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}< 0\)
\(\left[\begin{array}{nghiempt}\hept{\begin{cases}9-2\sqrt{x}>0\\\sqrt{x}-2< 0\end{array}\right.\\\hept{\begin{cases}9-2\sqrt{x}< 0\\\sqrt{x}-2>0\end{array}\right.\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}\hept{\begin{cases}x< \frac{81}{4}\\x< 4\end{array}\right.\\\hept{\begin{cases}x>\frac{81}{4}\\x>4\end{array}\right.\end{cases}\Rightarrow\left[\begin{array}{nghiempt}x< 4\\x>\frac{81}{4}\end{array}\right.}\)
Câu 3:
\(C=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
a) Để biểu thức \(\sqrt{\dfrac{2}{x^2}}\) xác định thì \(\left\{{}\begin{matrix}x^2\ge0\\x\ne0\end{matrix}\right.\)\(\Leftrightarrow x\ne0\)(vì x2>0)
Vậy x\(\ne0\) thì biểu thức \(\sqrt{\dfrac{2}{x^2}}\) xác định
b) Để biểu thức \(\sqrt{\dfrac{4}{x+3}}\) xác định thì \(x+3>0\Leftrightarrow x>-3\)
Vậy x>-3 thì biểu thức \(\sqrt{\dfrac{4}{x+3}}\) xác định
c) Ta có \(x^2>0\Leftrightarrow x^2+6>6\)
Mà -5<0
Suy ra \(\dfrac{-5}{x^2+6}< 0\)
Vậy với mọi x thì \(\sqrt{\dfrac{-5}{x^2+6}}\) sẽ không xác định
d) Ta có \(x^2\ge0\Leftrightarrow x^2+1\ge1\)
vậy \(\sqrt{1+x^2}\) sẽ xác định với mọi x\(\in R\)
a: ĐKXĐ: x>=0; x<>2
\(B=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Thay x=16 vào A, ta được:
\(A=\dfrac{16+3}{4-2}=\dfrac{19}{2}\)
Bài 1:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
b: \(P=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{x-3\sqrt{x}}{x-2\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
c: Để \(P=\dfrac{1}{2}\) thì \(2\sqrt{x}-6=\sqrt{x}-2\)
hay x=16
a) Để M xác định thì \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b) \(M=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
mơn nha