a,

A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99 

=1−(3+5+7+...+99)=1−(3+5+7+...+99)

=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498

b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98

c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99

ai giúp đc thì giúp đi!!! ai giúp mk cũng tick hết á!!!

\(-2\left(x-\frac{1}{2}\right)-5\left(\frac{3}{10}-1\right)=7\)

\(-2\left(x-\frac{1}{2}\right)-5.\frac{-7}{10}=7\)

\(-2\left(x-\frac{1}{2}\right)-\frac{-7}{2}=7\)

\(-2\left(x-\frac{1}{2}\right)=7+\frac{-7}{2}\)

\(-2\left(x+\frac{1}{2}\right)=\frac{7}{2}\)

\(x+\frac{1}{2}=\frac{7}{2}:-2\)

\(x+\frac{1}{2}=\frac{-7}{4}\)

\(x=\frac{-7}{4}-\frac{1}{2}\)

\(x=-\frac{9}{4}\)

\(-2\left(x-\frac{1}{2}\right)-5\left(\frac{3}{10}-1\right)=7\)

\(-2x+1-15+5-7=0\)

\(-2x-16=0\)

\(-2\left(x+8\right)=0\)

\(=>x+8=0\)

\(=>x=-8\)

= (11+9+2)+(1/2-3/2+5/2)-(2/3+5/3-7/3)

=22+1,5-0

=23.5

=(11+9+2)+(1/2-3/2+5/2)-(2/3+5/3-7/3)

=22+1,5-0

=23.5

Chúc học tốt

A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)

A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)

A=3.(1-1/400)

A=3.399/400

A=1197/400

A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)

A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)

A=3.(1-1/400)

A=3.399/400

A=1197/400

Bài 1: Tính nhanh:

A = 3/1*2 + 3/2*3 + 3/3*4 + ... + 3/399*400

=>3A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/399*400

    3A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/399 - 1/400

    3A = 1 - 1/400

      3A = 400/400 - 1/400

      3A = 399/400

        A = 399/400 : 3

        A = 399/400 . 1/3

        A = 133/400.

Có gì ko hiểu bn ib mk nha.^^

\(A=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{399.400}\)

\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)

\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)

\(A=3.\left(1-\frac{1}{400}\right)\)

\(A=3.\frac{399}{400}\)

\(A=\frac{1197}{400}\)

\(B=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{399.400}\)

\(B=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)

\(B=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)

\(B=5.\left(1-\frac{1}{400}\right)\)

\(B=5.\frac{399}{400}\)

\(B=\frac{399}{80}\)

\(C=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\)

\(C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\)

\(C=\frac{1}{5}-\frac{1}{151}\)

\(C=\frac{146}{755}\)

\(D=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{149.151}\)

\(D=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\right)\)

\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\right)\)

\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{151}\right)\)

\(D=\frac{3}{2}.\frac{146}{755}\)

\(D=\frac{219}{755}\)

\(E=\frac{11}{1.3}+\frac{11}{3.5}+\frac{11}{5.7}+...+\frac{11}{99.101}\)

\(E=\frac{11}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(E=\frac{11}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(E=\frac{11}{2}.\left(1-\frac{1}{101}\right)\)

\(E=\frac{11}{2}.\frac{100}{101}\)

\(E=\frac{550}{101}\)

_Chúc bạn học tốt_

\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{2}{5}\)

nhưng tại sao lại >1/2*3+1/3*4+1/4*5+...+1/9*10

bạn giúp mk nha