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Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
Ta có :
a , \(M=2\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(M=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}-\dfrac{2\left(x+9\right)}{x-9}\right]:\left[\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(M=\left(\dfrac{2x-6\sqrt{x}-2x-18}{x-9}\right).\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\right]\)
\(M=\dfrac{-6\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(2\sqrt{x}+4\right)}\)
\(M=\dfrac{-6\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(M=-\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b , mik ko chắc chắn nên mik chưa làm nhé !
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a) P=\(\left(\dfrac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\dfrac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)\(P=\left(1+\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}+\sqrt{x}-\sqrt{x}\right)\)
\(P=\left(1+2\sqrt{x}+x\right)\left(1-2\sqrt{x}+\sqrt{x}\right)\)
\(P=\left(1+x\right)^2-4x\)
\(P=1+2x+x^2-4x\)
\(P=1-2x+x^2\)
\(P=x^2-2x+1\)
\(P=\left(x-1\right)^2\)
b) Thay P=4 vào biểu thức ta có:
\(4=\left(x-1\right)^2\)
\(\Leftrightarrow\left(x-1\right)^2=2^2\)
\(\Leftrightarrow x-1=\pm2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy \(x_1=3;x_2=-1\)
a) P = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
P = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
P = \(\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
P = \(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
P = \(\dfrac{x-1}{\sqrt{x}}\)
b) Để P > 0
\(\Rightarrow\dfrac{x-1}{\sqrt{x}}>0\)
\(x-1>0\)
\(x>1\)
c) Để P = 6
\(\Rightarrow\dfrac{x-1}{\sqrt{x}}=6\)
\(x-1=6\sqrt{x}\)
\(x-6\sqrt{x}-1=0\)
\(\left(\sqrt{x}-3-\sqrt{10}\right)\left(\sqrt{x}-3+\sqrt{10}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=3+\sqrt{10}\\\sqrt{x}=3-\sqrt{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=19+6\sqrt{10}\\x=19-6\sqrt{10}\end{matrix}\right.\)
Bài 1:
\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)
Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Bài 2:
\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
a)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b)
\(Q< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}< 0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\ \Leftrightarrow0< x< 4\)
\(M=\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) với x>0;x≠1
\(=\left(\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\)
\(M=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{x}=\dfrac{-x+\sqrt{x}+2}{x\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)}{x\left(\sqrt{x}+1\right)}=\dfrac{2-\sqrt{x}}{x}\)
vậy M=\(\dfrac{2-\sqrt{x}}{x}\)
vì x>0 nên để \(M< 0\Leftrightarrow\dfrac{2-\sqrt{x}}{x}< 0\Leftrightarrow2-\sqrt{x}< 0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)