1/2 + 1/3 +....+1/63 >2
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\(\Leftrightarrow\dfrac{4}{9}:x=\dfrac{8}{3}\)
hay \(x=\dfrac{4}{9}\cdot\dfrac{3}{8}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)
1/2+1/3+1/4+.....+1/63>1/31+.....1/31(62 số hạng 1/31)
hay 1/2+1/3+1/4+.......+1/63>62x1/31
nên 1/2+1/3+1/4+......+1/63>2
Đặt \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{31}+...+\dfrac{1}{31}\)(có 62 số hạng \(\dfrac{1}{31}\))
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>\dfrac{1}{31}\times62\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\)
\(Vậy\) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}>2\left(đpcm\right)\)
1/2+1/3+1/4+...+1/63>1/31+1/31+1/31...+1/31( 62 số hạng 31)
hay 1/2+1/3+1/4+...+1/63>62 x 1/31
nên 1/2+1/3+1/4+...+1/63>2(dpcm)
k ủng hộ nha
c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)
\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)
\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)
\(=\dfrac{44}{75}\)