Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=3\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{1}{2ab}\ge\frac{3.4}{a^2+b^2+2ab}+\frac{2}{\left(a+b\right)^2}=\frac{14}{\left(a+b\right)^2}=14\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
Xét dạng tổng quát :
\(\sqrt{1+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}}=\sqrt{\frac{k^2+1}{k^2}+\frac{1}{\left(k+1\right)^2}}\)
\(=\sqrt{\frac{\left(k^2+1\right)\left(k+1\right)^2+k^2}{k^2\left(k+1\right)^2}}=\sqrt{\frac{k^4+2k^3+3k^2+2k+1}{k^2\left(k+1\right)^2}}\)
\(=\sqrt{\frac{\left(k^2+k+1\right)^2}{k^2\left(k+1\right)^2}}=\frac{k^2+k+1}{k\left(k+1\right)}=1+\frac{1}{k\left(k+1\right)}=1+\frac{1}{k}-\frac{1}{k+1}\)
Áp dụng vào bài toán :
\(A=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2010^2}+\frac{1}{2011^2}}\)
\(A=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2010}-\frac{1}{2011}\)
\(A=2009-\frac{1}{2011}+\frac{1}{2}\)
p/s: không biết tính có đúng ko nữa, bạn nhớ check lại. Mình nhớ bài này còn có cách khác ngắn hơn nhưng quên rồi :D
1) Để ý rằng : \(x\sqrt{x}-1=\sqrt{x^3}-\sqrt{1^3}=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)
\(P=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
2) \(x=28-6\sqrt{3}=\left(3\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=3\sqrt{3}-1\)
Thay vào P ta được :
\(P=\frac{3\sqrt{3}-1}{28-6\sqrt{3}+3\sqrt{3}-1+1}\)
\(P=\frac{3\sqrt{3}-1}{28-3\sqrt{3}}\)
3) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)
\(\Leftrightarrow x+\sqrt{x}+1>3\sqrt{x}\)
\(\Leftrightarrow x-2\sqrt{x}+1>0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)
BĐT cuối luôn đúng \(\forall x>1\)
Ta có đpcm
4) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{2}{7}\)
\(\Leftrightarrow2x+2\sqrt{x}+2=7\sqrt{x}\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy...
5) \(P=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(\Leftrightarrow Px+P\sqrt{x}+P=\sqrt{x}\)
\(\Leftrightarrow x\cdot P+\sqrt{x}\left(P-1\right)+P=0\)
Phương trình trên có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(P-1\right)^2-4P^2\ge0\)
\(\Leftrightarrow P^2-2P+1-4P^2\ge0\)
\(\Leftrightarrow-3P^2-2P+1\ge0\)
\(\Leftrightarrow-3\left(P^2+\frac{2}{3}P-\frac{1}{3}\right)\ge0\)
\(\Leftrightarrow P^2+\frac{2}{3}P-\frac{1}{3}\le0\)
\(\Leftrightarrow P^2+2\cdot P\cdot\frac{1}{3}+\frac{1}{9}-\frac{4}{9}\le0\)
\(\Leftrightarrow\left(P+\frac{1}{3}\right)^2\le\left(\frac{2}{3}\right)^2\)
\(\Leftrightarrow P+\frac{1}{3}\le\frac{2}{3}\)
\(\Leftrightarrow P\le\frac{1}{3}\)
Vậy \(maxP=\frac{1}{3}\Leftrightarrow x=1\)??
Đoạn này sai sai ta ?
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}<\frac{1}{5}.13=\frac{13}{5}<2\)