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A = \(\frac{4024x\left(2010+4\right)-2}{2011+2012x2010}\)= \(\frac{2024x2010+4024-2}{2011+2012x2010}\)=\(\frac{4024x2010+4022}{2011+2012x2010}\)= 2
câu B hình như sai đề bài . mk moi hoc lop 6 thoi nen cũng ko chắc .
mình không biết kq =mấy
nhứng mình c/m kq =2 là sai
\(A-2=\dfrac{4024.2014-2}{Khongquantam}-2=\dfrac{4024.2014-2-2.2011-2.2012.2010}{Khongquantam}\)
\(A-2=\dfrac{2\left(2012.2014-2011-2012.2010-1\right)}{Khongquantam}=\dfrac{2\left[2012.\left(2014-2010\right)-2011-1\right]}{Khongquantam}\)
\(A-2=\dfrac{2\left[4.2012-2011-1\right]}{Khongquantam}=\dfrac{2\left[3.2011+3\right]}{Khongquantam}\)
\(A-2=\dfrac{2\left[3.\left(2011+1\right)\right]}{Khongquantam}=\dfrac{2.3.2012}{Khongquantam}\ne0\)\(A-2\ne0\)
\(\Rightarrow A\ne2\Rightarrow kq=2=sai\)
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
=>x+2013=0
hay x=-2013
\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)
Ta có:
\(a^{2010}+b^{2010}+a^{2012}+b^{2012}\)
\(=\left(a^{2010}+a^{2012}\right)+\left(b^{2010}+b^{2012}\right)\ge2a^{2011}+2b^{2011}\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}a^{2010}=a^{2012}\\b^{2010}=b^{2012}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}\)
\(\Rightarrow a^{2013}+b^{2013}=2\)
Vậy \(S=2\)
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)
\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)
\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)
\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)
mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)
Vậy x=-2010
Tìm x:
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\dfrac{x}{2010}+\dfrac{x+1}{2011}+\dfrac{x+2}{2012}+\dfrac{x+3}{2013}+\dfrac{x+4}{2014}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2010}-1\right)+\left(\dfrac{x+1}{2011}-1\right)+\left(\dfrac{x+2}{2012}-1\right)+\left(\dfrac{x+3}{2013}-1\right)+\left(\dfrac{x+4}{2014}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{2010}+\dfrac{x-2010}{2011}+\dfrac{x-2010}{2012}+\dfrac{x-2010}{2013}+\dfrac{x-2010}{2014}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)=0\)
\(\Leftrightarrow x=2010\)
A=2.002983591
B= -1