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22 tháng 4 2020

Bài 1 : 

Ta có  : 

\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)

\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)

\(+\left(\frac{x+2013}{2011}+1\right)\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)

\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)

\(\Rightarrow x+4024=0\)

\(\Rightarrow x=-4024\)

22 tháng 4 2020

Bài 2 : 

Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)

=> Phương trình trở thành 

\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)

\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)

\(\Rightarrow5a^2+3a-8=0\)

\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)

Vì \(a\ge0\Rightarrow a=1\)

\(\Rightarrow x^2+2x+1=1\)

\(x^2+2x=0\)

\(\Rightarrow x\left(x+2\right)=0\)

\(\Rightarrow x\in\left\{-2,0\right\}\)

26 tháng 11 2019

Áp dụng BĐT Cô - si ngược dấu :

\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4\left(x-2010\right)}\le\frac{4+\left(x-2010\right)}{4}\)

\(\Rightarrow\sqrt{x-2010}-1\le\frac{4+\left(x-2010\right)}{4}-1=\frac{x-2010}{4}\)

\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}\le\frac{1}{4}\)

Hoàn toàn tương tự với những phân thức còn lại 

\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2010=4\\x-2011=4\\z-2012=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=2014\\y=2015\\z=2016\end{cases}}}\)

11 tháng 12 2019

\(A=\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}\)

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

\(=|1-x|+|x+2|\ge|1-x+x+2|=3\)

11 tháng 12 2019

\(x\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

\(\Leftrightarrow x\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\Leftrightarrow x\sqrt{x+\frac{1}{4}}=\frac{3}{2}\)

Làm nốt

7 tháng 5 2018

Ta có : 

\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=2012\)

\(\Leftrightarrow\)\(\frac{x-1-2012}{2012}+\frac{x-2-2011}{2011}+\frac{x-3-2010}{2010}+...+\frac{x-2012-1}{1}=0\)

\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\ne0\)

Nên \(x-2013=0\)

\(\Leftrightarrow\)\(x=2013\)

Vậy \(x=2013\)

Chúc bạn học tốt ~ 

7 tháng 5 2018

\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1+2012=2012\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

\(\Leftrightarrow x=2013\)

28 tháng 5 2016

Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)

28 tháng 5 2016

chóng váng

28 tháng 12 2015

\(N=\sqrt{1+2011^2+\frac{2011^2}{2012^2}+\frac{2012.2011}{ }}kolàsốtựnhieen\)

xem lại đề

30 tháng 12 2015

s k phai la stn dc

 

24 tháng 10 2020

\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)

Ta có \(f\left(x\right)+f\left(1-x\right)=1\) khi đó

\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)

\(=1+1+...+1+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\frac{1}{2}+3.\left(\frac{1}{2}\right)^2}=1005+\frac{1}{2}=\frac{2011}{2}\)

24 tháng 10 2020

Ta có: \(F\left(x\right)=\frac{x^3}{1-3x+3x^2}\)

\(\Leftrightarrow F\left(1-x\right)=1-\frac{x^3}{1-3x+3x^2}\)

\(=\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)

\(=\frac{\left(1-x\right)^3}{1-3x+3x^2}\)

Ta có: \(F\left(x\right)+F\left(1-x\right)\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3x+3x^2}\)

\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)

\(\Leftrightarrow F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)=1\)

...

\(F\left(\frac{1005}{2012}\right)+F\left(\frac{1007}{2012}\right)=1\)

Do đó: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)

\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)

\(=1+1+...+F\left(\frac{1}{2}\right)\)

\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)

\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)

\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)

\(=1005+\frac{1}{2}=\frac{2011}{2}\)

20 tháng 10 2016

Ta gán : \(1992\rightarrow D\)\(1992\rightarrow A\)

\(D=D+1:A=D.\sqrt[D]{A}\)

CALC , bấm liên tiếp dấu "=" cho đến khi D = 2013 thì dừng.

Sau đó bấm \(\frac{Ans}{D}\) sẽ ra kết quả cần tính.

8 tháng 10 2016

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