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\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+2016\right)\)
\(A=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+\frac{1}{4}.\frac{\left(1+4\right).4}{2}+...+\frac{1}{16}.\frac{\left(1+16\right).16}{2}\)
\(A=1+\frac{1}{2}.\frac{3.2}{2}+\frac{1}{3}.\frac{4.3}{2}+\frac{1}{4}.\frac{5.4}{2}+...+\frac{1}{16}.\frac{17.16}{2}\)
\(A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(A=\frac{1}{2}.\left(2+3+4+5+...+17\right)\)
\(A=\frac{1}{2}.\frac{\left(2+17\right).16}{2}=19.4=76\)
hik như vế sau là a làm theo 16 chứ k fai 2016 hay sao ấy
. Ta có: \(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\) \(\Leftrightarrow\frac{x+1}{2016}+1+\frac{x+3}{2014}+1=\frac{x+5}{2012}+1\frac{x+7}{2010}+1\)
. \(\Leftrightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\) \(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)\)
\(\Leftrightarrow x+2017=0\) \(\Leftrightarrow x=-2017\)
\(\frac{x+1}{2016}+\frac{x+3}{2014}=\frac{x+5}{2012}+\frac{x+7}{2010}\)
\(\Rightarrow\left(\frac{x+1}{2016}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+7}{2010}+1\right)\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}=\frac{x+2017}{2012}+\frac{x+2017}{2010}\)
\(\Rightarrow\frac{x+2017}{2016}+\frac{x+2017}{2014}-\frac{x+2017}{2012}-\frac{x+2017}{2010}=0\)
\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2017=0\)\(\left(Vì\frac{1}{2016}+\frac{1}{2014}-\frac{1}{2012}-\frac{1}{2010}\ne0\right)\)
\(\Rightarrow x=0-2017\)
\(\Rightarrow x=-2017\)
Vậy x=-2017
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
Ta có
\(M=a+\frac{2a+b}{2-b}+\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)
\(=a-\frac{2a+b}{b-2}+\frac{2a-b}{2+b}+\frac{4a}{\left(b-2\right)\left(b+2\right)}\)
\(=\frac{a\left(b-2\right)\left(2+b\right)-\left(2a+b\right)\left(2+b\right)+\left(2a-b\right)\left(b-2\right)+4a}{\left(b-2\right)\left(2+b\right)}\)
\(=\frac{ab^2-4a-4a-2ab-2b-b^2+2ab-4a-b^2+2b+4a}{\left(b-2\right)\left(2+b\right)}\)
\(=\frac{ab^2-8a-b^2}{\left(b-c\right)\left(b+2\right)}\)
Với \(b=\frac{a}{a+1}\)ta có
\(=\frac{a\cdot\frac{a^2}{a^2+2a+1}-8a-\frac{a^2}{a^2+2a+1}}{\left(\frac{a}{a+1}-2\right)\left(\frac{a}{a+1}+2\right)}\)
\(\frac{a\cdot\frac{a^2}{a^2+2a+1}-8a-\frac{a^2}{a^2+2a+1}}{\left(\frac{-a-1}{a+1}\right)\left(\frac{3a+1}{a+1}\right)}\)
\(=\frac{a\cdot\frac{a^2}{a^2+2a+1}-8a-\frac{a^2}{a^2+2a+1}}{\frac{1-3a}{a+1}}\)
\(=\frac{a\left(\frac{a^2}{a^2+2a+1}-8-\frac{a}{a^2+2a+1}\right)}{\frac{1-3a}{a+1}}\)
\(=\frac{a\left(\frac{-7a^2+15a+8}{a^2+2a+1}\right)}{\frac{1-3a}{a+1}}\)
tới đây tịt rồi ai làm tiếp đc k