Tinh : 1 - 1/2 : 1/3
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Ta có: \(4\left(1+\frac{\sqrt{3}}{2}\right)=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\Rightarrow1+\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}+1}{2}\right)^2\)
Tương tự \(1-\frac{\sqrt{3}}{2}=\left(\frac{\sqrt{3}-1}{2}\right)^2\)
\(VT=\frac{\left(\frac{\sqrt{3}+1}{2}\right)^2}{1+\frac{\sqrt{3}+1}{2}}+\frac{\left(\frac{\sqrt{3}-1}{2}\right)^2}{1-\frac{\sqrt{3}-1}{2}}=\frac{\frac{\left(\sqrt{3}+1\right)^2}{4}}{\frac{3+\sqrt{3}}{2}}+\frac{\frac{\left(\sqrt{3}-1\right)^2}{4}}{\frac{3-\sqrt{3}}{2}}\)\(=\frac{\left(\sqrt{3}+1\right)^2}{2.\sqrt{3}\left(\sqrt{3}+1\right)}+\frac{\left(\sqrt{3}-1\right)^2}{2.\sqrt{3}\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+1}{2\sqrt{3}}+\frac{\sqrt{3}-1}{2\sqrt{3}}=\frac{\sqrt{3}+1+\sqrt{3}-1}{2\sqrt{3}}=1=VP\)
các bạn ơi, giải nhanh nhanh cho mình nha. cảm ơn nhiều
(1-1/1+2)(1-1/1+2+3).....(1-1/1+2+3+....+2014)
= 1 + 2 / (2x3) + 2 : (3x4) + ... x 2 : (2014 : 2015)
= 1 + 2 x [ 1/(2x3) + 1/(3x4) + ... 1/(2014 x 2015)
= 1 + 2 x (1/2 - 1/3 + 1/3 - 1/4 - ... 1/2014 + 1/2014 - 1/2015 )
= 1 + 2 x ( 1/2 - 1/2015 )
= 1 + 1 - 2/2015
= 2 - 2/2015
= 4028/2015
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{20}.\left(1+2+....+20\right)\)
\(=1+\frac{1}{2}\times\frac{2.3}{2}+\frac{1}{3}\times\frac{3.4}{2}+...+\frac{1}{20}\times\frac{20.21}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)
\(=\frac{\left(2+21\right).20:2}{2}=\frac{230}{2}=115\)
Số cuối là
\(\frac{1}{10}.\left(1+2+3+...+10\right)\) hay \(\frac{1}{20}.\left(1+2+3+...+20\right)\) ??
A = 4008 + \(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2004}\)
= 4008 + \(\dfrac{1}{2.3:2}+\dfrac{1}{3.4:2}+...+\dfrac{1}{2004.2005:2}\)
= 4008 + \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2004.2005}\)
= 4008 + \(2\left(\dfrac{1}{2}-\dfrac{1}{2005}\right)\)
= 4008 + \(\dfrac{2003}{2005}\)
= 4008\(\dfrac{2003}{2005}\)
a ban oi , day la 4008 chia cho (1+1/1+2+1/1+2+3+.......+1/1+2+....+2004)