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a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
Giúp mk nhanh nha.mk cần gấp...........ai nhanh mak đúng mk k cho
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0
a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)
\(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow C=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3}{3-\sqrt{x}}\cdot\frac{x-3\sqrt{x}}{2\sqrt{x}+4}\)
\(\Leftrightarrow C=\frac{-3}{2\sqrt{x}+4}\)
b) Để \(-\frac{3}{2\sqrt{x}+4}< -1\)
\(\Leftrightarrow\frac{1+2\sqrt{x}}{2\sqrt{x}+4}< 0\)
Vì \(\hept{\begin{cases}1+2\sqrt{x}>0\\2\sqrt{x}+4>0\end{cases}\Leftrightarrow C>0}\)
Vậy để C <-1 <=> \(x\in\varnothing\)
c) \(A=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow A^2=3+2+2\sqrt{5}=5+2\sqrt{5}\)
\(B=\sqrt{5}+1\)
\(\Leftrightarrow B^2=5+1+2\sqrt{5}=6+2\sqrt{5}\)
Vì \(5+2\sqrt{5}< 6+2\sqrt{5}\)
\(\Leftrightarrow A^2< B^2\)
\(\Leftrightarrow A< B\)
Vậy \(\frac{1}{\sqrt{3}-\sqrt{2}}< \sqrt{5}+1\)
\(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)(ĐK: \(x\ge0,x\ne1\))
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A=\frac{5}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{5}{\sqrt{x}}\)
\(\Rightarrow x=5\left(x+\sqrt{x}+1\right)\)
\(\Leftrightarrow4x+5\sqrt{x}+1=0\)(vô nghiệm do \(x\ge0\))
\(A-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)(vì \(x\ne1\))
Do đó \(A< \frac{1}{3}\).