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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
Bài 1:
a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)
b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)
c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
a. \(\sqrt{\dfrac{63y^3}{7y}}\)=\(\sqrt{9y^2}\)=3y
b.\(\sqrt{\dfrac{48x^3}{3x^5}}\)=\(\sqrt{16\cdot\dfrac{1}{X^2}}\)= \(\sqrt{16}\cdot\sqrt{\dfrac{1}{X^2}}\)=\(4\cdot\dfrac{1}{X}=\dfrac{4}{X}\)
c.\(\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{\sqrt{9n^2}}{\sqrt{4}}=\dfrac{3n}{2}\)
d. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)
a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3y\)
b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{4}{x}\)
c) \(\dfrac{\sqrt{45mn^2}}{\sqrt{20m}}=\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{3n}{2}\)
d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)
a: \(S=\dfrac{x+1}{\sqrt{x}}:\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(x+1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
b: Khi \(x=\dfrac{2}{2+\sqrt{3}}=4-2\sqrt{3}\) vào S, ta được:
\(S=\dfrac{\left(4-2\sqrt{3}+1\right)\left(\sqrt{3}-1+1\right)}{\left(\sqrt{3}-1\right)\left(4-2\sqrt{3}-1\right)}\)
\(=\dfrac{\left(5-2\sqrt{3}\right)\cdot\sqrt{3}}{\left(\sqrt{3}-1\right)\left(3-2\sqrt{3}\right)}\)
P=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-\sqrt{x}}\right):\left(\dfrac{2}{x-1}+\dfrac{1}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{2}{x-1}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(a)P=\left(\dfrac{1}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+2\sqrt{x}+1}\\ P=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\\ P=\dfrac{1-\sqrt{x}}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ P=\dfrac{1-x}{x}\)\(b)P>\dfrac{1}{2}\Rightarrow\dfrac{1-x}{x}>\dfrac{1}{2}\\ \Leftrightarrow\dfrac{1-x}{x}-\dfrac{1}{2}>0\\ \Leftrightarrow\dfrac{2-3x}{2x}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2-3x>0\\2x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2-3x< 0\\2x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{2}{3}\\x>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{2}{3}\\x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{2}{3}\\x>0\end{matrix}\right.\)
M=(\(\dfrac{\sqrt{x}}{\sqrt{x}-x}-\dfrac{\sqrt{x}+2}{1-x}=\dfrac{\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}-\dfrac{\sqrt{x}+2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\))
M = \(\left(\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)-\dfrac{\left(\sqrt{x}+2\right)\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\sqrt{x}}\)
M=\(\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)-\left(\sqrt{x}+2\right)\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)=\(\dfrac{\sqrt{x}+x-x-2\sqrt{x}}{\sqrt{x}\left(1-x\right)}\)
M=\(\dfrac{\sqrt{x}-2\sqrt{x}}{\sqrt{x}\left(1-x\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}\left(1-x\right)}=\dfrac{-1}{1-x}\)
M= \(\dfrac{-1}{1-x}\) có giá trị nguyên khi 1-x là ước của -1
Các ước của -1 là :
1-x=1 suy ra x=0(loại)
1-x= -1 suy ra x=2