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Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{3\sqrt{x}-1}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x^3}-1}-\dfrac{2x+1}{\sqrt{x^3}-1}\right)\)
\(M=\left(\dfrac{x+\sqrt{x}-3\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{x-\sqrt{x}-2x-1}{\sqrt{x^3}-1}\right)\)
\(M=\left(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}+1}\right)\left(\dfrac{-x-\sqrt{x}-1}{\sqrt{x^3}-1}\right)\)
\(M=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\dfrac{-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(M=\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{2}{x-1}\right):\dfrac{2\sqrt{x}+4}{\sqrt{x}-1}\)
\(=\left(\dfrac{x+\sqrt{x}-x+\sqrt{x}+2}{x-1}\right)\cdot\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}+2\right)}=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
Câu 1:
a: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
=>a>4
\(A=\left(\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{\sqrt{x}}{1+\sqrt{x}}\right)+\dfrac{3\sqrt{x}}{x-1}\\ =\left(\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\dfrac{3\sqrt{x}}{x-1}\\ =\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}+\dfrac{3\sqrt{x}}{x-1}\\ =\dfrac{2\sqrt{x}-3\sqrt{x}}{1-x}=\dfrac{\sqrt{x}}{x-1}\)
\(A=\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{\sqrt{x}}{1+\sqrt{x}}+\dfrac{3\sqrt{x}}{x-1}=\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)+\sqrt{x}\left(1-\sqrt{x}\right)-3\sqrt{x}}{1-x}=\dfrac{-\sqrt{x}}{1-x}=\dfrac{\sqrt{x}}{x-1}\)
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
Đặt \(P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(P-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=-\dfrac{x-2\sqrt{x}+1}{x+\sqrt{x}+1}=-\dfrac{\left(\sqrt{x}-1\right)^2}{x+\sqrt{x}+1}\le0;\forall x\ge0\)
\(\Rightarrow P\le\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=1\) ko thỏa mãn ĐKXĐ nên \(P< \dfrac{1}{3}\)