M = \(\dfrac{5}{2.4}\) + \(\dfrac{5}{4.6}\)+ \(\dfrac{5}{6.8}\)+ ...+ \(\dfrac{5}{96.98}\)+ \(\dfrac{5}{98.100}\)

M = \(\dfrac{5}{2}\).( \(\dfrac{2}{2.4}\) + \(\dfrac{2}{4.6}\)+ \(\dfrac{2}{6.8}\)+...+ \(\dfrac{2}{96.98}\)+ \(\dfrac{2}{98.100}\))

M = \(\dfrac{5}{2}\).( \(\dfrac{1}{2}-\dfrac{1}{4}\)+ \(\dfrac{1}{4}-\dfrac{1}{6}\)+ \(\dfrac{1}{6}\) - \(\dfrac{1}{8}\)+...+ \(\dfrac{1}{96}\)-\(\dfrac{1}{98}\)+ \(\dfrac{1}{98}\)-\(\dfrac{1}{100}\))

M = \(\dfrac{5}{2}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\))

M = \(\dfrac{49}{40}\)

\(x\) \(\times\) M - 1 = \(\dfrac{20}{29}\)

\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{20}{29}\) + 1

\(x\) \(\times\) \(\dfrac{49}{40}\) = \(\dfrac{49}{29}\)

\(x\)           = \(\dfrac{49}{29}\) : \(\dfrac{49}{40}\)

\(x\)           = \(\dfrac{40}{29}\)

\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)

= \(\frac{5}{2}-\frac{5}{100}\)

= \(\frac{49}{50}\)

\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

    \(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

    \(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)

\(\Rightarrow Q=\frac{49}{40}\)

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)

\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)

\(A=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(A=\frac{5}{2}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(A=\frac{5}{2}.\left(1-\frac{1}{100}\right)\)

\(A=\frac{5}{2}.\frac{99}{100}=\frac{99}{40}\)

1/1×2+1/2×3+1/3×4+...+1/45×46+1/46+47

mình k viết lại đề nhé =)

câu A

A :5 =1/2.4+1/4.6+1/6.8+..+1/98.100

A:5 =1/2-1/4+1/4-1/6+1/6-1/8+...+1/98-1/100

A:5 =1/2-1/100 =49/100

A=49/100 x5 =49/20

câu B tươg tự nha =)

Ta có:

A =5/2(1/2-1/4 + 1/4-1/6+ 1/6..........1/98-1/100)

A =5/2 (1/2 -1/100)

A =5/2 x 49/100

A = 49/20

Xin lỗi: 5/2 x49/100 = 49/40 mới chuẩn

18:

a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)

=3*49/100=147/100

b: Để A là số nguyên thì n-1 thuộc Ư(2)

=>n-1 thuộc {1;-1;2;-2}

=>n thuộc {2;0;3;-1}

A=(1.2)(2.2)+(2.2)(3.2)+...+(50.2)(51.2)

A=1.2.4+2.3.4+...+50.51.4

A=4(1.2+2.3+...+50.51)

M= 1.2+2.3+...+50.51

3M=1.2.3+2.3.(4-1)+...+50.51.(52-49)

=1.2.3+2.3.4-1.2.3+...+50.51.52-49.50.51

= 50.51.52

=132600

=> M=44200

=> A=4M=176800

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)

\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)

                                                   \(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)

\(B=2.4+4.6+6.8+...+98.100\)

\(B=2.\left(1.2\right)+2.\left(2.3\right)+2.\left(3.4\right)+...+2.\left(49.50\right)\)

\(B=2\left(1.2+2.3+3.4+....+49.50\right)\)

Đặt:

\(A=1.2+2.3+3.4+...+49.50\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+49.50.\left(51-48\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50\)

\(3A=49.50.51\)

\(A=\dfrac{49.50.51}{3}=41650\)

\(B=2A=41650.2=83300\)