2/9,7/9,1 nửa,2/2,

Còn lại bó tay bởi vì em mới lên lớp 3 thôi à!

phan so  hay gi vay

Giải 

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

Số số hạng của A là: \(\left(100-1\right)\div1+1=100\)(số hạng)

Tổng A là: \(\frac{\left(100+1\right)\times100}{2}=5050\)

Vây A=5050

\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)

\(B=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)

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

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

Vậy \(B=\frac{99}{100}\)

minh cam thay de hoi sai

Rút gọn biểu thức trên nha.

\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)

\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)

vậy M=20

\(\text{-12(x-5)+7(3-x)=5 }\)

\(-12x+60+21-7x=5\)

\(-12x-7x=5-21-60\)

\(-19x=-76\)

\(x=-76:\left(-19\right)\)

\(x=4\)

\(\text{ 30(x+2)-6(x-5)-24x=100}\)

\(30x+60-6x+30-24x=100\)

\(30x-6x-24x=100-30-60\)

\(0=10\)

\(\Rightarrow x\)ko tồn tại

\(\text{(x+1-5)+7(3-x)=5 }\)

\(x+1-5+21-7x=5\)

\(x-7x=5-21+5-1\)

\(-6x=-12\)

\(x=\left(-12\right):\left(-6\right)\)

\(x=2\)

\(\text{(x+1)+(x+3)+(x+5)+........+(x+99)=0}\)

\(\text{(x+1)+(x+3)+...+(x+99)=0}\)
tổng các số hang là\(\frac{\left(99+1\right)}{2}=50\)(số hạng)
=>\(\text{(x+1)+(x+3)+...+(x+99)=0}\)<=> \(\text{50.x+(1+3+5+..+99)=0}\)
<=>\(\text{50.x+(99+1)}\)\(.\frac{50}{2}=0\)<=> \(\text{50.x+2500=0=}\)>\(x=\frac{-2500}{50}=-50\)

chúc bạn học tốt

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

Dãy trên có số số hạng là:

(100 - 1) + 1 = 100 (số hạng)

Tổng \(A=\frac{\left(100+1\right)\cdot100}{2}=5050\)

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

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

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

\(B=1-\frac{1}{100}\)

\(\Rightarrow B=\frac{99}{100}\)

~Học tốt~

giup mik di !!!!!!!

Nguyễn Đăng Duy ơi bài trên là tính nhanh hay tính vậy bạn .

A:tính số số hạng (100 số).

=>A=(1+100)*100:2=5050.

B=1/1*2+1/2*3+1/3*4+000+1/99*100.

=>B=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100.

=>B=1-1/100=99/100.

tk mk nha.đúng 1000% .

-chúc ai tk cho mk học giỏi và may mắn,thanks các bn nhìu-

a=100(100+1)/2

B=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100

B=1-1/100=99/100

Câu A tự làm nhé! Tính số số hạng rồi tính tổng

B = 1/1.2 + 1/2.3 + 1/3.4 +.....+ 1/99.100

B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +........+ 1/99 - 1/100

B = 1 - 1/100

B = 99/100

bạn ơi cho mk hỏi đề bài yêu cầu j thế ak

a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

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

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)

=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)

b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)

=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).

c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)

=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).

d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)

=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).

e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)