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25 tháng 7 2017

a. ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

\(B=\frac{2x+2}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}\)

b. Ta có \(B-5=\frac{2x+2\sqrt{x}+2}{\sqrt{x}}-5=\frac{2x-3\sqrt{x}+2}{\sqrt{x}}=\frac{2\left(x-2.\sqrt{x}.\frac{3}{4}+\frac{9}{16}\right)-\frac{9}{8}+2}{\sqrt{x}}\)

\(=\frac{2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}}{\sqrt{x}}\)

Ta thấy \(\hept{\begin{cases}2\left(\sqrt{x}-\frac{3}{4}\right)^2+\frac{7}{8}>0\\\sqrt{x}>0\forall x>0\end{cases}\Rightarrow B-5>0\Rightarrow B>5}\)

Vậy \(B>5\)

13 tháng 9 2019

\(C=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)

\(=\sqrt{x}-1\)

Ta co:

\(\sqrt{x}-1+\frac{2}{\sqrt{x}}=\frac{x-\sqrt{x}+2}{\sqrt{x}}=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}{\sqrt{x}}>0\)

\(\Rightarrow\sqrt{x}-1>-\frac{2}{\sqrt{x}}\)

23 tháng 6 2021

\(ĐKXĐ:x\ge0;x\ne1;0\)

\(A=\frac{2x+2}{\sqrt{x}}+\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(A=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

\(A=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)

\(A=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}\)

\(A=2\sqrt{x}+\frac{2}{\sqrt{x}}+2\)

a/d bđt cauchy 

\(2\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{2.2}=2.2=4\)

\(A\ge4+2=6\)

\(< =>A>5\)

dấu "=" xảy ra khi x=1