Question

x+30/100.x= -1,31

(4,5-2x).(\(-1\dfrac{4}{7}\))=\(\dfrac{11}{4}\)

A=1+2+3+4+5+...+99+100

B=\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)

Answer 1

1) \(x+\dfrac{30}{100}x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)

\(\Leftrightarrow100x+30x=-131\)

\(\Leftrightarrow130x=-131\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

Vậy \(x=-\dfrac{131}{130}\)

b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)

\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)

\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)

\(\Leftrightarrow-72+32x=77\)

\(\Leftrightarrow32x=77+72\)

\(\Leftrightarrow32x=149\)

\(\Leftrightarrow x=\dfrac{149}{32}\)

Vậy \(x=\dfrac{149}{32}\)

Answer 2

\(A=1+2+3+4+5+....+99+100\)

\(A=100(100+1):2=5050\)

\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+.....+\dfrac{1}{9900}\)

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)

\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(B=1-\dfrac{1}{100}=\dfrac{99}{100}\)