\(2\frac{1998}{1999}\)là hỗn số hay \(2.\frac{1998}{1999}\)hả bạn?

Là \(2.\frac{1998}{1999}\)

\(\sqrt{1.1998}< \frac{1+1998}{2}\)

\(S>\frac{2}{1999}+\frac{2}{1999}+...+\frac{2}{1999}=2.\frac{1998}{1999}\)

Áp dụng \(\frac{1}{\sqrt{a.b}}>\frac{2}{a+b}\) , ta có : 

\(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+...+\frac{1}{\sqrt{k\left(1998-k+1\right)}}+...+\frac{1}{\sqrt{1998.1}}>\)

\(>\frac{2}{1+1998}+\frac{2}{2+1997}+...+\frac{2}{k+1998-k+1}+...+\frac{2}{1998+1}=\)

\(=\frac{2.1998}{1999}\)

Vậy \(S>\frac{2.1998}{1999}\)

Sửa đề : \(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+...+\frac{1}{\sqrt{k\left(1998-k+1\right)}}+...+\frac{1}{\sqrt{1998.1}}\)

Tổng S có số số hạng là :(1998-1):1+1=1998(số)

Áp dụng bđt cosi vs hai số dương có

\(\sqrt{1.1998}\le\frac{1+1998}{2}=\frac{1999}{2}\)

\(\frac{1}{\sqrt{1.1998}}\ge\frac{2}{1999}\)

Tương tự cx có \(\frac{1}{\sqrt{2.1997}}\ge\frac{2}{1999}\)

..............

\(\frac{1}{\sqrt{k\left(1998-k+1\right)}}\ge\frac{2}{1999}\)

................

\(\frac{1}{\sqrt{1998.1}}\ge\frac{2}{1999}\)

=> \(S\ge\frac{2}{1999}+\frac{2}{1999}+...+\frac{2}{1998}\)

<=> \(S\ge2.\frac{1998}{1999}\)

a) \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\left(đk:x>0\right)\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\dfrac{1-x}{2\sqrt{x}}\right)^2=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}.\dfrac{\left(x-1\right)^2}{4x}=\dfrac{-4\sqrt{x}\left(x-1\right)}{4x}=\dfrac{1-x}{\sqrt{x}}\)

b) \(P-\left(-2\sqrt{x}\right)=\dfrac{1-x}{\sqrt{x}}+2\sqrt{x}=\dfrac{1-x+2x}{\sqrt{x}}=\dfrac{1+x}{\sqrt{x}}>0\)

\(\Rightarrow P>-2\sqrt{x}\)

a, ĐK: \(x\ge0;x\ne1\)

\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)

\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)

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