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\(\sqrt{1.1998}< \frac{1+1998}{2}\)
\(S>\frac{2}{1999}+\frac{2}{1999}+...+\frac{2}{1999}=2.\frac{1998}{1999}\)
\(2\frac{1998}{1999}\)là hỗn số hay \(2.\frac{1998}{1999}\)hả bạn?
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: \(x\ge0;x\ne1\)
\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)
\(=-\dfrac{x-1}{\sqrt{x}}\)
Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
\(\Leftrightarrow\frac{1}{\left(k+1\right)\sqrt{k}}-2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 0\)
\(\Leftrightarrow\frac{1-2k-2+2\sqrt{k\left(k+1\right)}}{\sqrt{k}\left(k+1\right)}< 0\)
Lại có: \(k>0\)
\(\Rightarrow k+1>0\)
\(\Rightarrow\sqrt{k}\left(k+1\right)>0\)
\(\Rightarrow-1-2k+2\sqrt{k\left(k+1\right)}< 0\)
Áp dụng BĐT Cô-si ta có:
\(k+\left(k+1\right)\ge2\sqrt{k\left(k+1\right)}\)
\(\Leftrightarrow2k+1\ge2\sqrt{k\left(k+1\right)}\)
\(\Leftrightarrow2\sqrt{k\left(k+1\right)}-2k-1\le0\forall k>0\)
Vậy \(\frac{1}{\left(k+1\right)\sqrt{k}}< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(dkxd:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(x-2\sqrt{x}+1\right)\cdot2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2\cdot2}{\left(\sqrt{x}-1\right)^2\cdot\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
Xét: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2\)
\(=\dfrac{2}{x+\sqrt{x}+1}-\dfrac{2\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(=\dfrac{2-2x-2\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{-2x-2\sqrt{x}}{x+\sqrt{x}+1}\)
\(=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)
Với \(x\ge0;x\ne1\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{x}\ge0\\x+\sqrt{x}+1>0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\ge0\)
\(\Leftrightarrow\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\le0\)
\(\Rightarrow A-2\le0\Leftrightarrow A\le2\)
Vậy: \(A\le2\).
Áp dụng bđt \(\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\left(a\ne b;a,b>0\right)\)ta có:
\(\dfrac{1}{\sqrt{1.1998}}>\dfrac{2}{1+1998}=\dfrac{2}{1999}\)
\(\dfrac{1}{\sqrt{2.1997}}>\dfrac{2}{2+1997}=\dfrac{2}{1999}\)
...
\(\dfrac{1}{\sqrt{1998.1}}>\dfrac{2}{1998+1}=\dfrac{2}{1999}\)
Cộng vế với vế ta được P > \(2.\dfrac{1998}{1999}\)
cảm ơn bạn nhóc trùm nhiều