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\(S=\frac{2^{2013}}{2^{2013}+1}+\frac{2^{2012}}{2^{2012}+1}+....+\frac{1}{2^{2012}+1}+\frac{1}{2^{2013}+1}\)
=(\(\frac{2^{2013}}{2^{2013}+1}+\frac{1}{2^{2013}+1}\))+(\(\frac{2^{2012}}{2^{2012}+1}+\frac{1}{2^{2012}+1}\))+...+ \(\frac{1}{2}\) ( có 2013 dấu ngoặc )
= 1+ 1+.....+ \(\frac{1}{2}\) = 2013\(\frac{1}{2}\)
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
a,
\(\Leftrightarrow\sqrt{1-x}=\frac{x-1}{\sqrt{6-x}+\sqrt{-5-2x}}\)
\(\Leftrightarrow-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\\-\sqrt{1-x}=\sqrt{6-x}+\sqrt{-5-2x}\end{cases}}\)
b,tự nàm
c,
\(\Leftrightarrow64x^2-64x-64=64\sqrt{8x+1}\)
\(\Leftrightarrow\left(8x+1\right)^2=10\left(8x+1\right)+64\sqrt{8x+1}+55\)
đặt \(\sqrt{8x+1}=a\)
=>a4=10a2+64a+55
nhận thấy phương trình có dạng x4=ax2+bx+c
tìm số m sao cho b2-4(2m+a)(m2+c)=0
sau đó đưa về (x2+m)2=k2 với k là 1 số bất kì,sau đó giải ra
b)đk \(x\ge1\)
\(\sqrt{1+x^2+\frac{x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}=\sqrt{\frac{\left(x+1\right)^2+x^2.\left(x+1\right)^2+x^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\sqrt{\frac{x^4+2x^3+3x^2+2x+1}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\sqrt{\frac{\left(x^2+x+1\right)^2}{\left(x+1\right)^2}}+\frac{x}{x+1}\)
\(=\frac{x^2+x+1}{x+1}+\frac{x}{x+1}=x+1\)
\(\Rightarrow\sqrt{1+2012^2+\frac{2012^2}{2013^2}}+\frac{2012}{2013}=2013\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2013\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2013\)
\(\Leftrightarrow x+\left|x-2\right|=2014\)
giai 2 pt
pt1 x+x-2=2014
x=1008
pt2 x+2-x=2014(vô lý)
Ta có : \(\left(x+\sqrt{x^2+2017}\right)\left(-x+\sqrt{x^2+2017}\right)=2017\left(1\right)\)
\(\left(y+\sqrt{y^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\left(2\right)\)
nhân theo vế của ( 1 ) ; ( 2 ) , ta có :
\(2017\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017^2\)
\(\Rightarrow\left(-x+\sqrt{x^2+2017}\right)\left(-y+\sqrt{y^2+2017}\right)=2017\)
rồi bạn nhân ra , kết hợp với việc nhân biểu thức ở phần trên xong cộng từng vế , cuối cùng ta đc :
\(xy+\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017\)
\(\Leftrightarrow\sqrt{\left(x^2+2017\right)\left(y^2+2017\right)}=2017-xy\)
\(\Leftrightarrow x^2y^2+2017\left(x^2+y^2\right)+2017^2=2017^2-2\cdot2017xy+x^2y^2\)
\(\Rightarrow x^2+y^2=-2xy\Rightarrow\left(x+y\right)^2=0\Rightarrow x=-y\)
A = 2017
( phần trên mk lười nên không nhân ra, bạn giúp mk nhân ra nha :) )
2/ \(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{4\sqrt{x-2011}-4}{x-2011}+\frac{4\sqrt{y-2012}-4}{y-2012}+\frac{4\sqrt{z-2013}-4}{z-2013}=3\)
\(\Leftrightarrow\left(1-\frac{4\sqrt{x-2011}-4}{x-2011}\right)+\left(1-\frac{4\sqrt{y-2012}-4}{y-2012}\right)+\left(1-\frac{4\sqrt{z-2013}-4}{z-2013}\right)=0\)
\(\Leftrightarrow\left(\frac{x-2011-4\sqrt{x-2011}+4}{x-2011}\right)+\left(\frac{y-2012-4\sqrt{y-2012}+4}{y-2012}\right)+\left(\frac{z-2013-4\sqrt{z-2013}+4}{z-2013}\right)=0\)
\(\Leftrightarrow\frac{\left(\sqrt{x-2011}-2\right)^2}{x-2011}+\frac{\left(\sqrt{y-2012}-2\right)^2}{y-2012}+\frac{\left(\sqrt{z-2013}-2\right)^2}{z-2013}=0\)
Dấu = xảy ra khi \(\sqrt{x-2011}=2;\sqrt{y-2012}=2;\sqrt{z-2013}=2\)
\(\Leftrightarrow x=2015;y=2016;z=2017\)
Đặt B là tên biểu thức
Với mọi n thuộc N*, ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) (*)
Áp dụng (*), ta được:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
Điều kiện \(\hept{\begin{cases}x-2011>0\\y-2012>0\\z-2013>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2011\\y>2012\\z>2013\end{cases}}}\)
\(\frac{\sqrt{x-2011}-1}{x-2011}+\frac{\sqrt{y-2012}-1}{y-2012}+\frac{\sqrt{z-2013}-1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{\sqrt{x-2011}}-\frac{1}{x-2011}+\frac{1}{\sqrt{y-2012}}-\frac{1}{y-2012}+\frac{1}{\sqrt{z-2013}}-\frac{1}{z-2013}=\frac{3}{4}\)
\(\Leftrightarrow\left(\frac{1}{x-2011}-\frac{1}{\sqrt{x-2011}}+\frac{1}{4}\right)+\left(\frac{1}{y-2012}-\frac{1}{\sqrt{y-2012}}+\frac{1}{4}\right)+\left(\frac{1}{z-2013}-\frac{1}{\sqrt{z-2013}}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{x-2011}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{y-2012}}-\frac{1}{4}\right)^2+\left(\frac{1}{\sqrt{z-2013}}-\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-2011}}=\frac{1}{4}\\\frac{1}{\sqrt{y-2012}}=\frac{1}{4}\\\frac{1}{\sqrt{z-2013}}=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2011=16\\y-2012=16\\z-2013=16\end{cases}\Leftrightarrow\hept{\begin{cases}x=2027\\y=2028\\z=2029\end{cases}}}\)
Ta có \(\frac{2012^{2013}}{2013^{2013}}=\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}\)
Vì \(\frac{2012}{2013}< 1\)nên\(\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}< \frac{2012^{2012}}{2013^{2012}}.1=\frac{2012^{2012}}{2013^{2012}}\)
hay \(\frac{2012^{2013}}{2013^{2013}}< \frac{2012^{2012}}{2013^{2012}}\)
\(\Rightarrow\frac{2012^{2013}}{2013^{2013}}+1< \frac{2012^{2012}}{2013^{2012}}+1\)
\(\Rightarrow\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}< \left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\)
Xét với n là số nguyên thì : \(\frac{1}{2^{-n}+1}+\frac{1}{2^n+1}=\frac{1}{\frac{1}{2^n}+1}+\frac{1}{2^n+1}=\frac{2^n}{2^n+1}+\frac{1}{2^n+1}=\frac{2^n+1}{2^n+1}=1\)
Vậy ta nhóm hợp lí như sau :
\(S=\left(\frac{1}{2^{-2013}+1}+\frac{1}{2^{2013}+1}\right)+\left(\frac{1}{2^{-2012}+1}+\frac{1}{2^{2012}+1}\right)+...+\left(\frac{1}{2^{-1}+1}+\frac{1}{2^1+1}\right)+\frac{1}{2^0+1}\)
\(=1+1+...+1+\frac{1}{2}\) (2013 số hạng 1)
\(=2013+\frac{1}{2}\)